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In 1872 Felix Klein, speaking at the University of ErJangen, 
suggested that various geometries should be distinguished by 
the groups of transformations under which their propositions 
remain valid. This exciting and important idea has had many 
repercussions in the world of mathematics and recently its 
effects have been felt in the school classroom: an outstanding 
feature of new mathematics syllabuses being the inclusion of an 
approach to geometry based on the study of such plane trans- 
formations as rotations and reflections. This study is doubly 
profitable, for not only do transformations help to throw 
geometric properties into sharp relief, but they also provide 
a fascinating introduction to group theory. Both of these 
aspects are given due consideration in Mr Jeger's book, which 
was described in Mathematics Teaching as 'perhaps the best 
development of school geometry from the group point of view 
which is to be found anywhere'. 

Readers new to this type of geometry will be surprised by the 
power and versatility of transformations and by the way that 
they can be used to solve many different types of construction 
problems in addition to such well-known results as the nine- 
point circle. The second aspect is of equal interest. The ways in 
which transformations are related and the groups they form 
are investigated. It is shown how the reflections generate ail the 
other congruence-preserving transformations, and that if, in 
addition, enlargements are considered, the result is a new group 
—the group of similarities which characterizes Euclidean 
geometry. Finally, a look is taken at the geometry associated 
with affine transformations. 

This book is very suitable for sixth-formers and under- 
graduates who want a not too abstract approach to groups, 
for teachers who want a compact and workable account of this 
kind of geometry, and for training college students as an intro- 
duction to new ideas in algebra and geometry. 

Jacket design by Timothy Drever 

Transformation Geometry 

Max Jeger 

George Allen & Unwin Ltd, 

Price in U.K. only: 25s. net 


This book is due for return on or before the 
last date shown above. 

A Series for Teachers and Students 


University of Leicester 

No. 1 


In the same series 
No. 2 


by A. W. Bell 

In preparation 
topology by J. E, Reeve 
vectors by E. H. Leaton 
logic by W. O, Storer 




Oberrealschule, Lucerne 



University of Southampton 








This book is copyright under the Berne Convention. 
Apart from any fair dealing for the purposes of 
private study, research, criticism or review, as 
permitted under the Copyright Act, 1956, no 
portion may be reproduced by any process without 
written permission. Enquiry should be made to 
the publisher. 

This translation © George Allen & Unwin Ltd., 1966 

Translated from the German 


© 1964 Raber Vcrlag Lucerne and Stuttgart 


in 10 point Times Roman type 



It is generally agreed that school mathematics syllabuses are 
in need of reform. The traditional syllabus is no longer an ade- 
quate preparation for mathematics as it is taught at a higher 
level ; it indicates very little of the range of contemporary uses 
of mathematics; and it contains a high proportion of routine 
computation and manipulation at the expense of mathematical 
ideas which yield immediate enjoyment and satisfaction. A 
number of schools are now experimenting with new syllabuses 
which attempt to cure these faults. 

Whether the experiments prove to be wholly successful or not, 
they are bringing a new element into the situation: an awareness 
that it is part of the job of the teacher of mathematics to inform 
himself about the relatively recent developments and changes in 
his subject. It is no longer possible to believe that developments 
in mathematics concern only the research mathematician and 
do not have any bearing on the mathematics taught in schools. 
This series of books is intended as a contribution to the reform 
of school mathematics by introducing to the reader some areas 
of mathematics which, broadly speaking, can be called modern, 
and which are beginning to have an influence on the content of 
school syllabuses. 

The series does not put forward explicit advice about what 
mathematics to teach and how it should be taught. It is meant to 
be useful to those teachers and students in training who want to 
know more mathematics so that they can begin to take part in 
the existing experimental schemes, or modify them, or devise 
their own syllabus revisions, however modest, The books are 
elementary without being trivial : the mathematical knowledge 
they assume is roughly that of a traditional grammar school 
course, although substantial sections of all the books can be 
understood with less. 

Now that the stability over a long period of school mathe- 
matics syllabuses seems to be corning to an end, it is to be hoped 
that a new orthodoxy does not succeed the old. The reform of 
mathematics teaching should be a continuing process, associ- 
ated with a deepening study of the subject throughout every 
teacher's professional life. These books may help to start some 
teachers on that course of study. d.w. 


The teaching of mathematics in grammar schools today is in 
an unsettled state. Transformation geometry, which has recently 
been introduced into school syllabuses, is the outcome of a 
series of reforms which have aimed at liberating the teaching of 
geometry from domination by Euclid. 

The traditional teaching of geometry is based on Euclid in 
two ways. First, logic always takes precedence with Euclid ; the 
arrangement of the material is essentially determined only by 
logic. Theorems are put next to each other if an abstract logical 
proof is possible, without regard to whether they belong 
together, or whether they are essential or unimportant. 
Secondly, the Euclidean style in teaching is distinguished by 
its emphasis on rigid figures; rigid congruence of triangles is 
considered to be the main method of proof in elementary 

The notion of transformation gives to modern teaching of 
geometry a central concept comparable to the idea of function 
in analysis. The hierarchy of groups of transformations leads 
to a natural order for the material, and instead of the static 
treatment of Euclid we have a dynamic geometry. 

Although transformation geometry dates back to Felix 
Klein (1849-1925), this reform is still fully in keeping with 
modern tendencies which demand that, in our teaching, more 
emphasis be placed on mathematical structures and isomor- 
phisms. Only in one respect can it be called old-fashioned — it 
seeks to preserve geometry as an essential part of mathematical 

The first two editions of this book produced a series of hints 
and proposals for furthering the use of transformation geometry 
in the classroom. Their aim was to make a larger circle of 
teachers familiar with these new ideas. The third edition has 
been completely revised. It has been extended into a course on 
transformation geometry which has been tried out several times 
during the last few years by the author in his own teaching. 
The general arrangement of the former editions, which built up 
theory and applications from concrete examples, has been 
preserved. The range of problems has been considerably 


Reforms will only prove acceptable if they leave open various 
ways of access from previous teaching. For this reason a too 
rigidly systematic representation has been avoided. This course 
presupposes some previous knowledge of geometry; it is 
assumed that the pupil knows the elementary parts of congru- 
ence and similarity geometry. 

Our book differs to some degree from previous school books 
on transformation geometry since it makes much more use of 
the group-theoretic structure of elementary geometry. Alge- 
braic operations for transformations are introduced at an early 
stage and are afterwards used extensively as a method of proof. 
We should like to throw open to discussion the question 'to 
what extent is this algebraic approach to school geometry 
likely to give the pupil more insight into the subject?' In the 
final sections an attempt has been made to build a bridge from 
constructive geometry to analytic geometry. 

The axiomatic system behind this course has, intentionally, 
not been discussed consistently. The author is convinced that in 
school teaching axiomatic discussions can at most have a local 
or retrospective character. But in the latter respect they belong 
to the advanced sixth form stage. 

The diagrams have been drawn by two pupils, A. Schenk and 
R. Ronchetti, from Lucerne. Dr. H. Loeffel has helped by 
reading the proofs. The author is most grateful to all these 


Lucerne, September 1963. 















Reflection in a Line 




Combination of translations 


Further example of groups 


4. Rotations 47 

Combination of rotations 

5. The Group of Isometries 66 

The role of reflections and half-turns in the group 
of isometries 

6. The Group of Transformations Mapping a 

Square onto itself 75 

7. Enlargements 82 
The group properties of enlargements 

Subgroups of the group of enlargements 

8. Similarities 98 
Direct similarities 

Opposite similarities 

9. Affine Transformations 1 10 
Properties of perspective affinities 

The group properties of perspective affinities 
Affine geometry 

10. The Affine Geometry of the Ellipse 132 



A bibliography of texts written in English on transformation 
geometry would hardly impress the casual reader by its length. 
Yet, although books on the subject are not numerous, there is 
already a wealth of notation and nomenclature available to 
budding authors. White preparing this version we have con- 
stantly been faced with the need to choose between alternative 
names for a particular transformation and the choice has not 
always been easy. For example, 'half-turn* is an admirable 
description of a rotation through 180°, but by adopting 
this name one loses the line-point analogy and, what is 
perhaps more important, a means of considering the trans- 
formation in a way that can be easily generalized to three 
dimensions. In this instance, as elsewhere, we have used the 
term which is most likely to have been encountered by school- 
teachers. The general similarities are not so well known as the 
isometries and here we have followed Coxeter in the use of 
'spiral similarity*. Since we have preferred 'enlargement' to 
'dilatation' (or 'dilation', for spelling provides an additional 
degree of freedom), we have not used 'dilative reflection' but 
have added 'stretch-reflection' to the geometer's vocabulary. 
We would plead in defence that the name is self-explanatory 
and can be usefully compared with 'glide-reflection'. 

One major change in notational practice has been made for 
this version. Textbooks recently published in England have 
tended to prefer the 'functional order* for the product of trans- 
formations, that is, TR is to be interpreted as first the rotation R 
and then the translation T. We have followed this convention 
which seems to have much to commend it. 

A.D., A.G.H. 


E(S,fi) an enlargement with centre S and scale factor [i 

H a half-turn about the point O 

M a a reflection in the line s 

H(0,9) a rotation about the point through the directed 

T a translation 

lV(S,0,fi) a spiral similarity with centre S, angle of rotation 
and scale factor (.i 

Z a stretch-reflection 

<&(s,z,ft) a perspective affinity with axis of affinity s, direction a 
and scale factor fi 

91 (A) the group of enlargements 

S (B) the group of translations and rotations 

93' the group of translations and half-turns 

S (K) the group of plane isometries 

Q (Q) the group of affine transformations 

Q a the group of perspective affinities with the given 

axis s 

S (S) the group of spiral similarities 

S* the group of similarities 

% (T) the group of translations 


We introduce the idea of a mapping by means of a simple 
example. Select one point S in the plane. For every point A we 
can find a point A' such that the line segment A A' has its 
mid-point at S. This condition ensures that a unique point A' is 
associated with A. Moreover, if A runs through all the points 
of the plane, then so does A'. Here we have a mapping of the 
plane onto itself. If we wish to give this relation a definite 
direction, then we call A the original, or object, point and A' the 
image point. 

If for every given object point A there is exactly one image 
point A', and if for every given point B' there is exactly one 
original B— that is, if the mapping is unique in both directions — 
then the mapping is called one-one. Our example has this 
property; here the inverse mapping, that is, the mapping which 
maps the image back onto the original point, is given by the 
same rule of construction. 

This example illustrates a very special type of mapping. The 
notion of mapping, however, includes every correspondence 
n 17 








Fig 2 



/ \ 

J. h , 

between sets of geometric objects. We shall give a rough outline 
of the range of possible mappings by means of a few further 

Let a and a' be two planes in space. A pencil of parallel 
lines (the lines being parallel neither to « nor to a'} produces a 
correspondence between the points of intersection with « and 
a'. This unique mapping is called a parallel projection from a 
onto a'. 

A pencil of parallel lines can also define a correspondence 
between the points of space and the points in a plane x. We 
associate with every point P in space the point of intersection of 
ji with the line passing through P. This is called a parallel 
projection of space onto the plane %. In this mapping every 
point P in space has a unique image P'\ however, there are 
infinitely many points which are mapped onto any particular 
image point Q\ The mapping is uniquely defined but is not 
one-one. If the parallel lines are perpendicular to the plane n, 
then the mapping is called the orthogonal projection of space 
onto the plane it (see Figure 2). 

The sets related by a mapping may contain different types of 
objects. For example, mappings between a set of points and a 
set of lines, or a set of points and a set of circles are considered 
in higher geometry.* When we draw plans and elevations of an 
object then we are mapping points in space onto a suitable set 
of triples of points in the plane. 

Similar correspondences between sets are also of great 
importance in other fields of mathematics. For example, the 

* Examples are poles and polars and cyclographic mappings. 



function/: x-*y defines a correspondence between the variables 
x and v. Let us consider y = x 2 . This defines a correspondence 
between the real numbers in the two intervals — 5<x<5 and 
0^y<25. It is easily verified that this correspondence is not 
one-one. However, the same function defines a one-one corres- 
pondence between the sets {x:l<x<5} and {j>:l<>><25}. If 
these sets of numbers are illustrated geometrically by point 
sets on two lines, we can consider the function to be a mapping. 
So we see that functions and mappings are related ideas. 

When 24 pupils sit in a classroom with 24 chairs, then we have 
a one-one correspondence between the set of pupils and the set 
of chairs. Here the correspondence is between finite sets. 
Mappings between finite sets also occur in geometry. For 
example, a rotation through 120° about the centre of an equi- 
lateral triangle ABC maps the vertices onto each other in the 
following way: A' = B, B' = C, C = A (see Figure 3). Here 
we have a mapping of the finite set {A,B,C} onto itself. One- 
one mappings of a finite set onto itself are called permutations. 

In this monograph we shall consider almost exclusively one- 
one mappings of the set of all points in the plane onto itself. 
We shall use the word transformation to describe this type of 

The transformations of the plane onto itself which are con- 
sidered in elementary geometry can all be based on construc- 
tions with compass and ruler. This course develops some topics 
of plane geometry using these transformations as a basis. 


Let s be some selected line in the plane. Let us now imagine that 
ail the points of the plane are traced onto a sheet of transparent 
paper. If we turn this sheet over in such a way that every 
point originally on ,v coincides with its original position, then 
we obtain a correspondence between points of the plane by 
transferring the points from the tracing paper back onto the 
plane. We call this mapping of the plane onto itself a reflection 
in the line s and we denote it by the symbol M s * We shall call s 
the axis of reflection. 

Reflection in a line is a transformation having the following 
obvious properties on which we shall base our further investiga- 

1. To every line s there corresponds a reflection A/,. If A' is the 
image of A under M„ then A is the image of A' (see Figure 
4(a)). We write 

M,{A) - A', MIA') = A. 

2. The points of s remain in the same position; they are the 
fixed points of the mapping M s . We say that s itself is a fixed 
line of the mapping. If A is not on s, then ^4' lies on the 
opposite side of s. The line AA' is also a fixed line ; it is, 
however, not pointwise fixed like s (see Figure 4(a)). 

3. The reflection M, preserves straight lines, that is, a straight 
line, g, has as its image another straight line, g' (see Figure 

* We here follow the current British notation; that is, we shall denote 
rotations by R and reflections by M (for mirror). 




.4 = 0' 

,F=F< A'=B, 


s A> 



Fig 4 


4. The reflection M s leaves distances and angles invariant. A 
closed polygon and its image, however, have opposite 
orientation (see Figure 4(b)). 

A particular consequence of 4 is that the angles and 0' 
shown in Figure 4(a) are equal. This implies that AA' and s are 
perpendicular lines. 

Together with the relation AF = A'F this gives a simple 
construction for obtaining images under the reflection M 3 : the 
line segment joining corresponding points A and A' is perpen- 
dicular to s and is bisected by it. 

5. If two different points A and A' are given arbitrarily, then 
there exists exactly one reflection M s such that M S (A) = A' 
and M ,(A') = A, The axis s is the mediator of the segment 

6. Given two rays FA and FB issuing from a point F, then there 
exists exactly one reflection M s interchanging the two rays. 
The axis s is the bisector of the angle AFB. 

Problem 1. Given s, a point A, A' = M S (A% and a point X, 
construct the image point A" using only a ruler (that is, only 
straight lines may be drawn). 

Hint: remember that if B = A', then B' = A. Every pair of 
corresponding points, therefore, defines a second pair. 



Problem 2. Given are a line s and two points A and B on the 
same side of 5. Find the point X on 5 for which the length of the 
path composed of the two line segments AX and XB is a 

Imagine the segment XB reflected in s. Its length is not 
changed, that is, we have 

AX+XB = AX+XB'. 

We can immediately find the minimum of AX+XB'; it is 
given by the segment AB'. The two parts AX and XB of the 
minimal path lie on symmetric lines g and g' (see Figure 5). The 
point X is characterized by the fact that the two lines intersect 
s in three equal angles. 

According to Fermat's principle, light follows the path for 
which the time of travel is a minimum. In a homogeneous 
medium minimum time of travel is equivalent to minimum 
distance travelled. Reflection of light in a plane mirror, there- 
fore, leads to Problem 2. Figure 5 provides a basis for the 
solution by construction of many problems on reflection. 

Problem 3. A ray of light issuing from a point A is to be 
reflected in two lines St and s 2 in such a way that it finally 
passes through a given point B (see Figure 6). 

The solution can be developed from that given for Problem 
2. Now, however, we must make two reflections. 

In the solution shown in Figure 6, A has first been reflected in 



Fig 7 

s t and its image A i mapped onto A 12 by reflection in the lines 2 . 
Another possible method of solution is shown in Figure 7. Here 
B has been mapped onto B 2 by reflection in the fine s 2 . In 
both constructions the line segments drawn (namely, A l2 B and 
A l B 2 ) have the same length as the path of light from A to B. It 
should be mentioned that while the problem always has a 
geometrical solution this does not mean that there is always a 
sensible physical solution. For example, for a solution to have 
any physical meaning the two points of reflection P and Q must 
lie above the intersection of the lines s, and s 2 . 

We now return to Figure 6. Reflection in s 1 maps g onto g y 
which is mapped onto g t2 by reflection in s 2 . We may take g 12 
as being obtained directly from g if we interpret the result of 
performing the two reflections, one after the other, as a new 
transformation. This expresses an essential idea of transforma- 
tion geometry : transformations can be combined. The result of 
this composition is a new transformation. 

In order to clarify this idea we now consider the combination 
of two reflections away from our particular construction prob- 
lem. We map the points of the plane first by the reflection M a in 
a line a, and then map the images obtained by a reflection M b . 
The correspondence between a point A and its final image 
A l2 = M b (M a (A)) defines a new transformation which we shall 
denote by M b M„. Note that this notation indicates the order of 
composition as 'first M a , then M b — as is usual with functional 
notation we read from right to left. Figure 8 illustrates the 
construction of this new mapping. 

The process of combining transformations just described is 



Reflection M a Reflection M b 

«-M a MJ- 

Combined Transformation M a M a 

■M b (M e (g)i 

not restricted to reflections in lines; it produces from two 
arbitrary transformations <& and *P a new transformation NKO. 
This symbol means that each point is first mapped by <D, then 
its image is mapped by V. 

Now we can see that the set of all mappings of the plane onto 
itself has a property we have already met in algebra. Combina- 
tion of transformations is an operation associating with any 
two transformations a new transformation, that is, an object of 
the same kind. We find a similar situation, for example, with 
the set of all positive numbers if we consider multiplication as 
the operation of composition. By stressing these analogies our 
geometric considerations will assume an algebraic aspect. 

If a reflection M, is performed twice, we obtain the mapping 

which leaves all points of the plane in the same position, that is, 
every point of the plane is a fixed point. This trivial mapping is 
called the identity transformation and it is denoted by the 
symbol /. In the set of all transformations, / plays a similar 
part to that of the number 1 in the set of numbers mentioned 
above ; for every transformation <& we have 



Definition. If we reverse the correspondence between points 
given by a transformation <5, we obtain the inverse transforma- 
tion to <D. This is denoted by <D~ l . 


It follows from this definition that, for every transformation 

<J»(I> ' = O '<!> = /'. 

Definition. Any transformation (other than the identity) which 
is the same as its inverse is called an inuolutory mapping or 

From d> _1 = <J> we deduce that an involution satisfies the 
equivalent relation 

O* = * 2 = /. 

Reflections in lines, therefore, are examples of involutions. 

Problem 4. Show by means of suitable drawings: 

(a) M p Mq is not, in general, the same transformation as 
A/,M p , that is, the combination of reflections is non-commuta- 

(b) M p M q = M t Mp if jp and q are orthogonal lines. 
Figure 9(a) illustrates the first assertion. 






Fig 9 


So as to prove the second assertion, we show that every point 
A is mapped by the combined transformation onto the same 
image irrespective of the order in which the reflections are 
carried out (see Figure 9(b)). 

»/fj — *A X ' 


M q M p 

M p M q 


From the invariance of distances and angles we deduce 

A,0, ^,'arecollinearand OAJ = OA\ ^ ^ , = A , _ j, 
A, O, A 2 ' are collinear and OA 2 ' = OAf ' * 2 

The mapping M f M q = M q M p is easily constructed if we note 
that is the mid-point of the line segment A A'. We call this 
transformation reflection in the point O or a half-turn about O 
and we denote it by H . 

From the relation 

H 2 m H H = M p M q M„M p - / 
we see that a half-turn is an involution.* 

* We have here assumed that the operation of combining transforma- 
tions is associative, that is, that {MpM^(MJtf^l = M„((M,M,)M S ). The 
reader should check the validity of this assumption for himself. 



In connection with this problem we want to mention briefly 
the question of parallels. The quadrangle AA 1 A'A 2 is obviously 
a rectangle. However, we can only prove this if we use some 
axiom which expresses, in some form or other, Euclid's postu- 
late on parallels. If we want to base plane geometry on the 
study of transformations, the following postulate appears to be 
especially suitable. 

If three angles in a quadrangle are right-angles, then so is the 
fourth angle. (Axiom of the existence of the rectangle.) 

From this we deduce that the quadrangle OPAxQ is a 
rectangle. The same is true for the three other quadrangles in 
corresponding positions. But then we have proved that all the 
angles in the quadrangle AA^A'A-l are right-angles. 

Problem 5. Show that for every rectangle there exist two 
reflections M p and M q which map the rectangle onto itself. The 
axes p and q are orthogonal. 

Definition. If a reflection M p maps a figure onto itself, then we 
say that p is an axis of symmetry of the figure. If the figure is 
mapped onto itself by a half-turn H , then we say that O is a 
centre of symmetry of the figure. 

Figure 10 shows two reflections which map the square ABCD 
onto itself, the axes of symmetry being s t and s 2 . The square 
has four axes of symmetry; a regular H-sided polygon is easily 
shown to have n axes of symmetry. For a circle every diameter 
is an axis of symmetry. 

Wallpaper patterns, especially friezes, supply us with many 
simple examples of figures with infinitely many axes or centres 



a — 


* > 



of symmetry. (We must here assume that the pattern is extended 

Definition. Two lines g and h are called parallel if both are 
orthogonal to a line a. 

It is easy to deduce from this that parallel lines are also 
characterized by the equality of a pair of alternating angles 
with a line a. 

Problem 6. Deduce from the definition that: 
/■|gandg||/j *>f\\h. 


Problem 7. We are given two parallel lines p and q and two 
points A and B between them. We wish to construct the path of 
a ray of light issuing from A which passes through B after being 
reflected twice in both p and q, the first reflection being in p. 
Let g be the line along which the light begins to travel. Then 
the subsequent parts of the path lie on the image lines of g under 
the mappings M p , M q M p , M p M q M p and M q M„M q M p . The 
problem can be solved by constructing the images of A under 
these transformations, namely A u A 2 , A 3 and A'. The line 
through A' and B is the required Imeg' and once this has been 
drawn the remaining parts of the path can be found easily. 
Note that we could save a little space by making use of the fact 
that M„ and M q are involutions. For 

M q M p M„M p {g) m g' 


and, therefore, 

M p M q {M q M p M q M p {g)) = M p M q (g'). 

But, using the involution property, 

M,M q M q M p M q M p (g) = M q M p (g). 

That is, we can obtain g 2 by considering the line joining 
A 2 = M q M p (A) to B 2 = M p M q (B). This obviates the need to 
construct A'. 

Problem 8. Solve Problem 7 if only three reflections are de- 
manded (for example, one in p and two in q). 

Problem 9. A rectangle has width p and length 2p. The sides of 
the rectangle lie on the lines a, b, c and d, and its centre is M. 
Construct the hexagon MA { BCDA 2 with A x and A z on a, B on 
b. C on c and D on d, which has the smallest perimeter subject 
to these conditions. Express this minimal perimeter in terms of 


We can solve the problem by considering the image of M 

under the transformation M e MiM c M b M a . 

However, we prefer to split this chain into the products 
M c M b M B and M a M d : this leads us to the two points M 3 
= M c M b M a {M) and M 2 ' = M 4 M a {M) (cf. Problem 7). The 



minimal perimeter is equal to the length of M 3 M 2 '. We find 
that it is 5p, 

If we consider the symmetry of this problem, we can see 
immediately that C is the centre of a side of the rectangle. If C 
is known the construction can be considerably simplified. 

In the solutions of Problems 3, 7, 8 and 9 we have met 
products of a finite number of reflections. These are transforma- 
tions leaving distances and angles unchanged. Hence, if a figure 
is mapped by a product of a finite number of reflections, then 
its image is congruent to the original. 

Definition. Finite products of reflections are called congruences 

or isometries. 

Fig 73 

Figure 1 3 represents a product of three reflections 

We can see from the figure that the inverse transformation is 
*~ l = M p M q M r . 

given by 

Theorem 1. We can find the inverse transformation to a finite 
product of reflections by reversing the order of composition of 
the reflections. 



After these remarks we turn to some further types of prob- 
lems whose solutions can be found by means of reflections. 

Problem 10. A line s and two circles T u T 2 are given. Construct 
squares with two opposite vertices on s and with one of the 
remaining vertices on each of the circles T, and F 2 . 


Assume that A and C lie on s. The vertices B and D are then 
symmetrically placed with respect to s t that is, M S (B) = D. 
Imagine all the squares that could be drawn with X and Zona 
and Y on T t . The locus of the vertices Y' would then be the 
image circle F/ (= MJtTx)). The vertex of the square to be 
constructed must, therefore, be a point of intersection of F t ' 
and F 2 . 

This solution is based on the reflection of a locus of points. 
This is a new way of using mappings to help us to solve con- 
struction problems. 

Problem 11. A line s and two circles r 1( T 2 on the same side of s 
are given. Construct a point X on s with the property that 
tangents from X to the two circles make equal angles with s. 



The condition means that the two tangents through X are 
symmetric with respect to the axis s, that is, M s maps one onto 
the other. Therefore, X must be the intersection of s with a 
common tangent to the circles r t and T 2 '. In general, therefore, 
the problem has four solutions. 

Problem 12. A line g and two points A, B on the same side of g 
are given. Find a point X on g such that the angle between XB 
and g is twice as large as that between XA and g. 


Fig 16 

We now consider the product of two reflections in parallel 
axes p and q (see Figure 16). 

M q M p maps A onto the point A' on the perpendicular to p 
and q through A; the distance AA' being twice the distance 
between p and?. If we consider this new transformation without 
reference to M p and M q , we can characterize it as follows : 

The Hues joining corresponding points P and P' are parallel. 

All points are moved the same distance in the same direction. 

Definition. A mapping satisfying the above conditions is called 
a translation. 

A translation is obviously determined if one arbitrary pair 
P, P'— consisting of a point and its image— are given. This pair 
determines a direction and a distance— two quantities which 
can be represented geometrically by an arrow leading from P to 
P'. The arrows associated with different pairs of corresponding 
c 33 



Fig 17 

points are all equivalent since each of them determines the same 

Definition. A class of equivalent arrows is associated with every 
translation. The equivalence class of arrows is called a vector. 
Every one of the arrows serves to represent the vector. 

We denote vectors in the text by small letters printed in bold 
type, for example, a, b, «, v, and in the figures by the symbol — 
placed below the letters. 

If a translation is determined by a vector v, we find the images 
of the points A, B, C, . . . by attaching the vector v to each of the 
points in turn (see Figure 17). 

We now return to the generation of translations by composi- 
tion of reflections (Figure 16). The translation resulting from 
the combination of M p and M q has a very simple relation to the 
lines/) and q: the translation vector, v, is twice as long as, and 
in the same direction as, a, the vector perpendicular top which 
runs from a point on p to a point on q. Here we have to note the 
order of composition; for the mapping M q M p the vector a 
points from p to q. We express the relation between v and a by 
the equation v = 2a. 

We now consider a translation T with vector v. If p and q 
are two parallel lines such that the vector a, which is perpendi- 
cular to p and runs from p to q, satisfies the equation a = ^r, 
then T = M q M p . Since there are infinitely many pairs of parallel 
lines which satisfy these conditions we have the following 



Theorem 2. A translation T with vector v can be represented in 
infinitely many ways as a product of two reflections. The axes 
of the two reflections are parallel, are orthogonal to v, and are 
a distance apart equal to half the length of p. 

We shall use this theorem to help us to analyse isometries. 
It is obvious that translations, being products of reflections, 
are isometries. We deduce the following properties. 

1. Translations are one-one mappings. 

2. Translations are line-preserving transformations; moreover, 
the image g' of a line g is always parallel to g. (See Figure 

3. Translations are direct isometries, that is, congruences which 
preserve orientation. 

Fig 19 


The inverse transformation to a translation Twith vector v is 
again a translation. Its vector has the same length as v, but 
opposite direction; it is denoted by — v (see Figure 19). 

Before investigating translations further we shall consider 
two elementary constructions based on translation. 

Problem 13. Two lines g and h are given. Construct a line x 
making an angle of 60° with g and such that its points of inter- 
section with g and h are a distance d apart. 



Fig 20 

The angle of 60° with g and the length d determine a vector v 
(see Figure 20). A translation with vector v maps g onto g'. 
The required line x passes through the intersection of g' and It 
and is parallel to v. 

The problem has four solutions; it is easy to see that, apart 
from v, the vectors w, - v and -w will all lead to lines with the 
necessary properties. 

Problem 14. Given are two circles T, and T 2 and a \ineg. Find 
a point A on Tj and a point B on T 2 such that A and B are a 
distance </ apart and A3 is parallel to g. 

Here a solution is obtained by mapping one of the circles by 
means of a translation with a vector a which is parallel to g and 
has length d. 


When we consider the product of translations the half-turn, 
which was defined on p. 26, will prove to be an extremely useful 
aid and so we shall now consider this particular congruence in 
more detail. 

First we show that a translation can be represented as the 
product of two half-turns. For this purpose we split the trans- 
lation T into two reflections M p and M q (see Figure 21). If s is 
perpendicular to p and q, then 

T - M q M p = M q M,M s M p = (M t M s )(M s M P ) 

— fJ g H f , 

using the fact that the product of reflections in orthogonal lines 



Fig 21 

Fig 22 

is a half-turn about their point of intersection. 
We immediately deduce: 

Theorem 3. A translation T with vector v can be represented in 
infinitely many ways as a product of two half-turns. The 
centres of the half-turns F and G have to be chosen in such a 
way that FG = $v. 

Problem 15. Show that the product of three half-turns is again a 

Let the product be £1 = H„H G H F . 

We solve the problem by considering the translation 
H G H e = T, which has the vector v = 2FG. Choose the point K 
such that KH = FG (see Figure 22). Then we have 

« - H H {H G H F ) m HJPM = H K , 

which proves our assertion. 

Opposite sides of the quadrilateral FGHK are parallel and of 
equal length. This can be shown by multiplying 

H C H F = H,[H K 

by H F on the right and by H n on the left. We then obtain 

ffii^G^F^r ~ HnHtiHicHf <*■ H n H c = H K H F . 

But this implies GH = FK. Hence the points F,G,H,K are the 

vertices of a parallelogram. 

Remembering that a half-turn is an involution, we deduce the 

« = rif[Jtj G ii F HijM G H F = /. 


Problem 16. Find the image of the point A under ft 2 by finding 
successive images under half-turns about F,G,H,F f G,H- Of 
what geometric figure do these image points form the vertices ? 
We note that 

Theorem 4. In a product of three half-turns the order of the 
factors may be reversed. 

Problem 17. Let T be a translation with vector v and let H F be 
a half-turn about f. Show that TH F and H f Tzltc also half-turns 
and construct their centres from v and F. 

We now consider two translations 7\ and T 2 with vectors a 
and b respectively. We split each translation into two half- 

Fig 23 

Combining T t and T 2 we obtain the mapping 

r 2 7\ = H B H G H C H P = H H H Fy 

which is clearly again a translation, being the product of two 
half-turns. On the other hand 

T X T 2 = H G H F HffH G = {H g H f Hj^Hq 
= {HiiH F H G )H G = H B H F . 

Here we have used Theorem 4 which said that the order of 
factors in a product of three half-turns could be reversed. 

Comparing the two representations for T 2 Ti and T t T z we 

T t T 2 = T 2 Ti = T. 



Fig 2'i 


Theorem 5. The product of two translations is a translation. 
Moreover, the order of composition does not matter, that is, 
combination of translations is commutative. 

We associate with the new translation T the vector c. It can 
be obtained by considering the map of a single point A. The 
image of A under 7", is A* and T 2 maps A* onto A'. The vector 
c is, therefore, represented by an arrow from A to A'. We call 
c the sum of the two vectors a and b, and write 

c = a+b. 

This associates with the operation of combining translations 
an operation defined for vectors which is called vector addition. 

We obtain the vector sum of a and b by attaching the vector 
b to the 'point' of the vector a. The vector sum is then the vector 
which starts at the starting point of a and ends at the end-point 
of b (see Figure 24). This construction appears to depend 
upon the order in which the addition takes place, but this does 
not influence the result since 

T 2 Ti = T 1 T 2 *>a+b = b+a (see Figure 25). 




Fig 25 

Fig 26 

Translations and vectors provide us with two ways of 
expressing the same geometrical facts. The product T Z T % in the 
language of translations corresponds to the vector sum a+b. 

If we perform the same translation twice, we describe this by 

T = TiTi = T^o-c = a + a = la. 

The vector c has the same direction as a, but is twice as long 
(see Figure 26). 
If 7*i is performed n times, we have 

T = T t * o e = a+a+ ... +a = na. 

n times 

Here we have introduced the symbol na for the vector which 
has the same direction as a, but is n times as long. 

For later applications we shall need the following generaliza- 

Definition. Let / be a real number. We denote by Xa the vector 
with a length equal to the length of a multiplied by \X\, and 
which has the same or the opposite direction as a according to 
whether /. is positive or negative.* 

This definition includes the previously defined vector —a; 
this is the vector Xa for X = — 1. 

Certain geometric properties lead to further algebraic 
operations for vectors. The theory of these operations is called 
vector algebra. We are here only interested in vector addition 

•The definition uses implicitly the following principle of continuity: 
if a line segment is given on a line, then every other segment on the line can 
be characterized by a real number: its measure in relation to the given line 



and in multiplication of vectors by a number. A further study 
of vector algebra is not, therefore, necessary for our purposes. 
Vectors are of great importance in physics. Most physical 
quantities are vectors or similar to vectors. We mention as 
examples forces, velocities, accelerations, electric and magnetic 


We now consider the set £ of all translations. Included in 
this set is the identity translation which is interpreted as the 
translation with a vector of length zero.* We found that when 
two translations were combined the resulting transformation 
was again a translation (Theorem 5). Hence, when two ele- 
ments of % are combined they always yield another element of 
X, The operation of composition does not take us outside the set 
X It has further been shown that associated with every transla- 
tion there is an inverse transformation which is also a transla- 
tion. Every element ofX has an inverse which belongs to X. Such 
sets with an operation defined upon them having the above 
properties are extremely important in modern mathematics. 
They are known as groups. So as to enable us to develop this 
idea we now give a precise definition of a group. 

Definition. A set <S of objects (elements) A,B,C, 
group if the following postulates are satisfied. 

is called a 

I. An operation is defined on © which associates with every 
ordered pair A, B of elements of ® a new element C which 
also belongs to (5, C is called the product of A and B and we 

C = AoB. 

II. There exists an element E of t5 satisfying the relation 

AoE = EoA = A 

for all elements A of ®, 

£ is called the unit, or neutral, element of the group. 

• The direction of this vector is irrelevant. In vector algebra it is known 
as the zero vector. 



III. For every element A of <5 there exists an inverse element, 
denoted by A~ l , which satisfies 

AoA~ l m A~ l oA = E. 

IV. The operation is associative, that is, 

(AoB)oC m Ao(BoC). 

Thus translations form a group when we take the operation of 
composition to be the result of performing the two mappings 
one after the other. The unit element is the identity translation. 
The inverse element corresponding to a translation with vector 
v is the translation with vector — v. Postulate IV is automatically 
satisfied by the operation of combining transformations. 

. t 




Fig 2? 

Further examples of groups of transformations may be 
obtained, for example, by considering only those translations 
which map a certain figure onto itself. Let us ask, for instance, 
which translations map the frieze pattern shown in Figure 27 
onto itself. (We imagine that the pattern has been extended 
indefinitely in both directions.) One of them is certainly the 
translation with the vector v. But the translations with the 
vectors 2v, 3p, — p, — 2v, , . . also leave the figure unchanged. 
The figure, in fact, is left unchanged by any translation with a 
vector of the form Xv, where X is an arbitrary integer. It is easy 
to verify that the set of all such translations forms a group. We 
obtain the group of all translations which map the given linear 
pattern onto itself. 

The parquet pattern shown in Figure 28 is also left invariant 
under certain translations provided we think of the pattern as 
being continued indefinitely. If two such translations are 
combined, then we obtain another mapping of the pattern onto 



itself. The translations with vectors v lt v 2f P3, v 4 are examples 
of transformations which map the pattern onto itself. It is easily 
seen that all the translations with this property have vectors of 
the form 

V sa Ap,+jif 2j 

where X and ft are integers, for example, p 3 = 2v l + v 1 , 
p 4 = p 1 +2p 2 . These mappings again form a group; we call it 
trie group of translations of the given plane pattern. 

The idea of a group is not restricted to transformations. The 
definition of a group applies to other sets and operations. The 
study of such systems has developed into a special branch of 
mathematics— group theory. This theory provides a common 
framework for statements and methods concerning the most 
diverse fields of mathematics, provided that they are based on 
the same logical structure. Here one encounters mathematical 
thinking in its purest form. 

In connection with the two groups of translations of patterns 
we should like to remark that groups of mappings of figures are 
sometimes used to characterize works of art, patterns and types 
of crystals.* 

•See, for example, Coxeter, Introduction to Geometry, pp. 35, 278-9; 
Terpstra, Some notes on the Mathematical Background of Repetitive 
Patterns; Weyl, Symmetry. 



The group of translations which leave the linear pattern of 
Figure 27 invariant contains only transformations which are 
also elements of the group of all translations. Here we have a 
subset of the full group of translations which is itself a group. 
One talks in this context of a subgroup. 

Definition. A subset S of a group © is called a subgroup of (5 if 
the elements of § themselves form a group under the operation 
defined on (5. 

The group of translations which leave the plane pattern of 
Figure 28 invariant is also a subgroup of the group of all 

Problem 18. A square A BCD is given. Let a be the vector with 
starting point D and end-point B, and let b be the vector from 
D to C. Find the image of the square under the translation with 
the vector 2a +36. 

Problem 19. In a square ABCD we define vectors u = AB 
and v = AD. What is the figure formed by the images of this 
square iT it is mapped by all the translations having vectors 
-?.B+;u* where A and /* are integers? 


1. The group of integers under the operation of ordinary 
addition. The unit element is the number and the inverse 
element to the number n is —n. 

2. The group of even integers under the operation of addition. 
This is a subgroup of the previous example. 

3. The group of rational numbers under the operation of 
addition. This group contains the two previous examples as 

4. The group of non-zero rational numbers under the opera- 
lion of ordinary multiplication. Here the unit element is the 
number 1 and the inverse clement to n is l/«. 

5. The group of positive rational numbers under the operation 
of multiplication. This is a subgroup of the group of 
Example 4. 



Let E be the symbol for an even number and O the symbol 
for an odd number. We take ordinary addition of numbers 
to be our operation of composition. Then we obtain the 
following rules of addition : 

E+E = E (even + even = even) 

£+ = (even + odd = odd, independent of order) 

O + O = E (odd + odd = even). 

This defines the operation of addition on the set {0,E} and 
we can easily verify that these two elements form a group 
under the operation + . When a group contains only a finite 
number of elements, then we can describe it by means of a 
group table such as that below. 


+ ! E O 

First E 

O E 

1. We divide all the integers into three classes according to the 
remainder left when they arc divided by 3. We use the symbol 
R for all integers divisible by 3, R^ for those which leave 
the remainder 1 and R 2 f° r those integers which leave the 
remainder 2. There are no other integers. If, once more, we 
use addition as our operation, we obtain the following 

Rq+Rq = Rq, 

R l + R i = R 2 , 

Ro + R^ = Ri, 

R 1 + R z m R , 

Rq + R 2 = ^2' 

R 2 + R 2 — Ri- 

We can easily check that this system forms a group with 
three elements. The group table has the following form : 


R Q R t R 2 


Rq Ri R 2 


/?! R 2 Rq 

R 2 

R 2 Rq Ri 


The three symbols R , Ry and R 2 represent the residue 
classes modulo 3 ; this group is, therefore, called the group of 
residue classes modulo 3. Here R is the unit element: the 
inverse element to S u for example, is R 2 . 

Problem 20. Check whether or not the following sets form 
groups under the given operations. 

(a) The set (5 of ali positive fractions of the form l//i, under 


(b) The set SR of all residue classes modulo 5 other than the 
zero class (that is, the set 91 = {R u R 2 , R 3 , R*}), under 

(c) The set SR of all residue classes modulo 6 other than the 
zero class, tinder multiplication. 

(d) The set K of all integers, under subtraction. 

(e) The set $ = {a+by/2:a,b rational; a 1 +b z ^0}, under 

{/)The set of all translations with vectors of the form Xa+pb 
where a and b are given vectors and X, ji are irrational num- 
bers, under the operation of combination of transforma- 

We conclude this first introduction to groups with an example 
of a non-associadve operation. Let ty be the set of all points in 
the plane. With each pair of points A and B in ^3 we associate 
the point C which is the mid-point of the line segment AB. The 
reader should verify that the operation so defined is not asso- 
ciative. This example shows that one must not always assume 
the associativity of an operation. 


Fig 29 

We now consider the product of two reflections M p and M q 
whose axes, p and q, intersect in the point O. The transforma- 
tion R = M q M p is a congruence which preserves orientation. 
Such congruences are known as direct isometries. The position 
of corresponding points A and A' = R(A) is such that 
OA = OA' and the angle 

= LAO A' 

has a fixed value. We have in fact 

L.AOA' = 2« 1 +2« 2 = 2<t», 

where <b is the angle between the axes of reflection p and q 
(see Figure 29). 

The mapping R is called a rotation about O through the angle 
= 2<j>. 6 is known as the angle of rotation and O is the centre 
of rotation. 

Problem 21. Verify the relation /_AOA' = 2<f> for different 

positions of A. 




A rotation R is obviously determined once the centre and 
angle and direction of rotation are known. Angle and direction 
of rotation can be considered together as one quantity known 
as a directed angle. In this respect it is usual to consider angles 
to be positive if the direction of rotation is anticlockwise. Our 
statement concerning the product of M p and M q can now be 

Theorem 6. If p and q are two lines which intersect at O, then 
M q M p is a rotation about and the angle of rotation is twice 
the directed angle between p and q. 

At the same time we can deduce the following theorem : 

Theorem 7. A given rotation R about a point O and through a 
directed angle can be expressed in infinitely many ways as the 
product of two reflections. The axes of reflection p and q can 
be any pair of lines through O such that <f>, the directed angle 
between p and q, satisfies the relation 

Using Theorem 7 all rotations can be reduced to a product of 
reflections in exactly the same way as we found to be the case 
for translations. This will prove to be of value when we come to 
analyse isometrics. 

We can immediately deduce the following properties of 




Fig 30 

1. A rotation is a one-one mapping; the inverse mapping to 
R(0,8) is the rotation ^"'(0,-0). 

2. Rotations are line-preserving transformations. The angle 
between a line g and its image g' is equal to the angle of 

This property is easily verified from Figure 30. The line h 
is the perpendicular to g through O and h' is its image. The 
angle between g and g' is y = 0. 

3. A rotation has, in general, exactly one fixed point, namely 
the centre of rotation 0. The exceptional cases arise when 8 
is a multiple of 360° in which case every point is a fixed point, 
that is, we have the identity transformation. 

We shall now familiarize ourselves with this new type of 
transformation by making use of rotations to solve some 

construction problems. 

Problem 22. Rotate a square A BCD about a given point 
through an angle 8 = —60°. 

Problem 23. A point A and two lines b and d are given. Con- 
struct a square ABCD such that B lies on b and D on d* 

We begin the solution of this problem by ignoring the 
condition on D. The remaining conditions determine an easily 
identifiable family of squares. If we draw the square with 
vertex X on b, then the opposite vertex X' is the image of X 
under a rotation about A through an angle 8 = 90° (see Figure 
3 1). If X moves on b, then X' moves on the image of b under 
this rotation, that is, on the line b'. 

The problem has two solutions, since the rotation through 
8 = - 90° will also provide a solution. 

The solution to Problem 23 depended on the mapping of a 
geometric locus by a rotation. This same idea can be used to 
solve Problems 24 to 30. 

* We assume that the vertices of the square are A, B, C, O in that order, 
cf. Problem 95. 



Fig 31 

Problem 24. Construct an equilateral triangle which has one 
vertex at A and one vertex on each of the given lines b and c. 

Problem 25. A point A and two lines b and c are given. Con- 
struct an isosceles triangle ABC such that B lies on b, C lies on 
c and I ABC = LACB = 75°. 

Problem 26. Given are a point A and two concentric circles 
r, and r 2 . Construct an equilateral triangle with one vertex at 
A, the vertex B on T x and the vertex C on T 2 . 

Problem 27, Given are a point P and two circles T ( and r 2 . 
Find a point „4 on T l and a point 5 on T 2 such that /> is the 
mid-point of AB. 

Problem 28. Given are two circles Tj and T 2 intersecting in the 
points P and Q. Construct a line through P which intersects 
the two circles in chords of equal length. 

For the solution of Problems 27 and 28 we use rotations 
through 180°, that is, half-turns. 


Problem 29. Given are three points A, B and C. Construct a 
square with centre A such that two adjoining sides (or their 
extensions) pass through B and C respectively. 

Problem 30. P and Q are two points inside a circle F. Construct 
two chords of T of equal length, which intersect at an angle of 
45° and such that one passes through P and the other through 

We now turn to another type of problem which can be solved 
by means of rotations. Now, however, the angle of rotation is 
not fixed ; it depends upon how we draw the initial figure. 

Problem 31. A circle T and a point P are given. Construct a 
second circle with a given radius r which passes through P and 
which has a common chord with T of length s. 

We first ignore the condition that the circle should pass 
through P. Then there is a family of circles satisfying the 
remaining conditions and we can easily construct one of them; 
the circle with centre M (see Figure 32) is a representative of 
this family. To obtain a solution to the problem this circle must 
be rotated about O until it passes through P. 

Problem 32. Given are a circle V and a line g. Construct a 
triangle with given sides which has the vertices A and Hour 
and the vertex C on g. 



Fig 33 

If we ignore the last condition we obtain two families of 
triangles, one for each orientation of the triangle. In Figure 33 
we have drawn one representative of each of the two families. 
A suitable rotation about O will bring C x or C 2 onto the line 3. 

Problem 33. Two concentric circles Fj and T z , and a line g are 
given. Construct an equilateral triangle having sides of given 
length, d, and one vertex on each of T lt T 2 and g. 

We now continue our study of the theory of rotations by 
considering yet another type of problem. 



Problem 34. AB and A'B' are two given directed line segments 
having the same length but not parallel. Show that there always 
exists a rotation R which maps A onto A' and B onto B'. 

The problem is solved if we can find two reflections M p and 
M„ such that M M p (A) = A', M q M„(B) = B'. 

We take p to be the mediator of AA' and let B* denote the 
image of B under M p . If we now take q to be the mediator of 
B*B', then M p and M 4 are the two reflections which satisfy our 
requirements. The centre of rotation (the point of intersec- 
tion of p and q) can also be found by a construction based on 
symmetry: O is the point of intersection of the mediators of 
A A' and BB' (see Figure 34). From this we can see that there is 
essentially only one rotation giving the required mapping (that 
is, there is only one rotation if we restrict the angle of rotation 
to lying between and 360°). 

Problem 35. What happens to the construction above if the 
line segments AB and A'B' are 

(a) symmetric with respect to a line, 

(b) parallel but in an opposite sense? 

Under the assumptions about AB and A'B' which are listed 
in Problem 34 we can always find a rotation having the required 
properties. It seems reasonable, therefore, to see what happens 
in the exceptional case when the two directed line segments are 

Assuming that AB and A'B' are parallel directed line seg- 
ments having equal lengths, then our construction leads to 
parallel axes of reflection. Hence, the transformation that 

Fig 35 



maps AB onto A'B' is a translation. Translations can be con- 
sidered as special cases of rotations, namely as rotations about 
a centre at an infinite distance through a zero angle. 

Fig 36 

Problem 36. A rectangular table top of length 2 yd. and of 
width 1 yd. is to be fixed to the supporting frame in such a way 
that, by turning about an axis perpendicular to its plane, it can 
take the two positions indicated in Figure 36. Is this possible, 
and if so, about which point must the table top rotate ? 

Using the labelling of vertices indicated in Figure 36, the two 
rectangles ABCD and A'B'C'D' are congruent and have the 
same orientation. Since corresponding sides are not parallel, 
there must be a rotation which maps one onto the other. The 
centre of rotation is the point of intersection of the mediators of 
line segments connecting corresponding points. If we take 
A, A' and M,M' as the pairs of corresponding points, we 
obtain mediators which are orthogonal and, moreover, parallel 
to the sides of the rectangle. The coordinates of O with respect 
to ABCD are then easily calculated. 

It should be noted that there are two essentially different ways 
of associating the vertices of the two rectangles. Consequently 
the problem has two solutions. In Figure 36 the angle of rotation 
is Q m +90°; for the other solution the angle is = -90°. 



We now consider the transformation resulting from the 
composition of two rotations. We shall see that, in general, we 
obtain another rotation and this will lead us to a new group of 

We start with: 

Problem 37. Map the square ABCD by the rotation 
Ri(Oy,0i = 60°) and then map its image by the rotation 
R 2 (O 2 ,0 2 = 90°). Show that the resultant mapping is again a 

Figure 37 shows the various steps of the construction. The 
mapping R 2 R t is obviously an isometry which preserves 
orientation, and it must again be a rotation. For, according to 
Problem 34, there is exactly one rotation which maps A D onto 
A'D'. This mapping, however, maps ABCD onto A'B'C'D'. 
The centre of rotation O and the angle of rotation are easily 
constructed from AB and A'B'. 

Fig 37 




Fig 3d 

We analyse the product of two rotations further by consider- 
ing a suitable decomposition of J?, and R 2 into reflections. Let 
s be the line through the two centres of rotation, O t and 2 . By 
Theorem 7, we can decompose J?! and J? 2 in infinitely many 
ways into two reflections. We now select two lines p and q such 

K, = M S M P , R 2 = Af,Af a . 

R = R 2 R V = M q M s M s M P = M,M r 

The fixed point of R is the point of intersection of/? and q : 
the angle of rotation is twice the directed angle between p and q. 
From Figure 38 we immediately deduce the relation 

= 0i+0 2 . 

Problem 38. Find, using reflections, the product of two rotations 
R t and R 2 when the angles of rotation 8 t and 9 2 have different 

If we take into account the signs of the angles we once more 
find that 

8 = 8 t + 9 2 (see Figure 39). 

Fig 39 



Theorem 8. When rotations are combined, the angles of rotation 
are added algebraically. 

A"fes: — 


Fig 40 

ft =-30 

$2= +30° 

Fig. 41 

If the two angles of rotation differ only in sign, that is, 
8y+8 2 = 0, then every line g is mapped onto a parallel line 
g' (see Figure 40). 

Figure 41 illustrates how this case is analysed using reflec- 
tions. We can see immediately that R 2 Ri = M t M p is a transla- 


Problem 39. Given are a translation with vector v and a rotation 
about the point O through an angle = 60°. The product of 
the two mappings, in the given order, is again a rotation through 
6 = 60°. Construct the centre of this rotation. 

Theorem 9. Rotations and translations together form a group S 
under the operation of combination of transformations. 

To prove this theorem we have to verify that all the group 
postulates are satisfied. This is easily done, using the results 
already obtained, and we leave it to the reader. 

If two rotations R t and R 2 are performed one after the other, 
the final position of a given figure will, as a rule, depend upon 
the order in which the rotations are performed. Rotations are 
not, in general, commutative. This is best shown by an example. 

Let us take R t to be a rotation about the point 0, through an 
angle Q t = — 60° and let R 2 be a rotation about 2 through 
0, « -90°. 

Consider the effect of the two mappings, one after the other, 
on a simple figure such as the line segment AB joining the two 
centres of rotation. The construction shows that the position 
of the final image of AB depends upon the order in which the 
two rotations are performed (see Figure 42). The mapping 
R 2 Ri (that is, first R t then R 2 ) maps AB onto A V 'B X '\ R X R 2 
maps AB onto A 2 'B 2 . 

The angle of rotation in both cases is t +0 2 . Consequently 
the two image line segments are parallel. 

The group of all translations and rotations, therefore, differs 
essentially from the group of all translations since the product of 
two of the group elements now depends upon the order in 
which the two elements are written. We say that the group is 
non-commutative. We note, however, that there are certain pairs 
of mappings in the group which do commute. For example, the 
product of two rotations about the same point is independent of 
the order in which the rotations are performed. However, the 
commutative law no longer holds in general. 

Problem 40. R t and R 2 are two rotations. Construct the centres 
of R 2 Ri and R l R 2 using suitable decompositions of R t and R 2 
into products of reflections. 




B,=B> 2 

Fig 42 

Fig. 43 

Let s be the line between the two centres, we can then use 
the following decomposition which is indicated in Figure 43: 

R t m M,M p = M,.M„ R 2 = M € M S = Af.jtf,. 

We obtain 

R 2 Rl = M q M p , R t R 2 m M p .M q .. 

The fixed points of the two rotations R 2 R X and RiR 2 are, 
therefore, symmetrically placed with respect to the line s. 

Problem 41. Given are two rotations R x and R 2 . Find those 
points which are mapped onto the same image point under i?, 
and R 2 . 
Hint: a point with this property is a fixed point under R 2 ~ 1 R 1 . 

Problem 42. If i? t and R 2 are two arbitrary rotations, show that 
the centres of R lt R 2 R t and R 2 ~ l Ri are collinear. 

Problem 43. A, B and C are the vertices of a triangle (described 
in a clockwise direction), and a, 0, y are the angles at A, B and 
C respectively. Show that the three rotations Rt(A, 2a), 
R 2 {B t 20), R 3 (C t 2y) satisfy the relation 

R$R 2 R) — I, 



Problem 44. Give pairs of elements of the group S which 

commute with each other. 

Problem 45. Given are four points A, B, Cand D. Find a square 
such that each side (or its extension) passes through one of the 
given points. 

We analyse this problem in the following way (see Figure 44). 
A rotation R about M, the centre of the square, through an 
angle = 90 s maps the square onto itself. In particular, the 
strip between a and c is mapped onto the strip between b and d. 
The vector u (from A to C) is mapped onto a vector w which 
has the same length as u and is at right-angles to u. The rotation 
cannot, of course, be realized immediately since M, the centre 
of rotation, is still unknown. However, let us consider the 
product of this rotation and a translation T with vector t. 
This maps C onto C* = D and maps A' onto A*. The vector 
h> is not changed since vectors are invariant under translations — 
it is only moved to another position. 

The mapping TR can now be constructed. It maps a onto the 

Fig. 44 



Fig. 45 

vector w which has the above mentioned properties and, more- 
over, has its end-point at D. We can now construct this vector 
without any difficulty. But once we have constructed w we have 
found A*, that is, a second point on the side of the square 
through B. This enables us to draw the side b and the remaining 
sides of the square can then be readily constructed. 

We leave it to the reader to check that this problem has six 
different solutions. Figure 45 shows how the square is con- 
structed for a special disposition of the points A, B, C, D. 

Problem 46. Given are three points A, B, C. Construct a square 
which has A as a vertex and such that one of its two sides not 
passing through A passes through B and the other through C. 
This is a special case of Problem 45. Here two of the given 
points coincide. 

In the solution of Problem 45 we used a rotation TR which 
mapped the strip between the lines a and c onto the strip 
bounded by b and d. In particular, a was mapped onto b and c 
onto d. There are infinitely many rotations through an angle 
8 = 90° which do this. We now consider the problem of finding 
the geometrical locus of the centres of such rotations. 

Problem 47. Given are two perpendicular lines g and g' which 
intersect at the point S. Find the geometric locus of the centres 
of all rotations which map g onto g' (see Figure 46). 





Every rotation mapping g onto g' maps a line segment PQ 
on ^ onto a line segment P'Q', of the same length, on g'. In 
drawing P' and Q' we have to take into account the direction of 
rotation. The rotation is completely determined by the choice of 
such a pair of line segments. The centre of rotation, O, is the 
intersection of the mediators of PP' and Q Q\ The fact that 
9 = 90° means that O lies on the circles having PP' and QQ' as 
diameters. These two circles, moreover, pass through S. Hence, 
we have 

LOSQ m lOQQ' = 45°. 

The geometric locus of the centres of all the rotations through 
an angle = 90° which map g onto g' is, therefore, one line 
bisecting the angle between g and g'. 

There are also rotations with 6 = — 90° which map g onto 
g' ; the geometric locus of their centres is the other bisector. 

Problem 48, This is a generalization of Problem 47. Given are 
two lines g and g' which now meet at an arbitrary angle iff. Find 
the geometric locus of the centres of all the rotations mapping 
g onto g'. 

Problem 49. Show that the set of all translations and half- 
turns forms a subgroup SB' of SB. 
The proof of this should present no particular difficulty. 



We denote the fact that the elements of SB' are also elements 
of SB by writing SB'c SB. 

If we again denote the set of all translations by I, then we 
have the following chain of inclusions: 


We conclude our study of the group SB' by considering some 
typical problems. 

Problem 50. Show that the product of an odd number of half- 
turns is always a half-turn. 

Hint: use the fact that the product of two half-turns is a 

Problem 51. Given five points P, Q, R, S, 7", construct a pentagon 
having these points as the mid-points of its sides. 

The vertices of the pentagon have been denoted by F, F lt F 2 , 
F 3 , F 4 (see Figure 47) Fis mapped onto itself by the following 
chain of half-turns: 





II p Hq 

H a 

>F 3 >F 4 >F' » F. 



Fis, therefore, a fixed point of the mapping 

CI = rliH^ti^MnMp^ 

From Problem 50, £1 is a half-turn. Its centre, F, can be con- 
structed by mapping any point A in the plane by CI. F is then 
the mid-point of A A'. 

The fixed point F can also be obtained without mapping a 
figure (cf. Problem 15 and Figure 22). 

We note that this problem always has a solution. 

Problem 52. Construct an n-sided polygon, when the mid- 
points Jl/ ( (i = 1, . . . , «) of all the sides are given. 

This is a generalization of Problem 51. Let us denote the 
half-turn about M t by H i and consider 

£2 = H„H„^ , . , . H 2 H l . 

n is a translation or half-turn according to whether n is even 
or odd. This leads to two completely different solutions. If n is 
odd, then the problem always has a unique solution just as in 
the case n = 5. 

For even «, however, the problem has a solution only if Q 
is the identity; otherwise CI has no fixed point. But if £1 = /, 
then every point in the plane is a fixed point; hence, in this 
case there are infinitely many u-sided polygons with the given 
points as mid-points of sides. 

Consider in the two simplest cases, « = 4 and n = 6, what 
condition is imposed on the given points by demanding that 
Cl = I. 

If we now apply Theorem 4 to the three left-hand factors we 

H D H E H C H C H B H A = / <*" H D H E H B Hji = I. 

This last relation holds if, and only if, A, B, E, D arc the 
vertices of a parallelogram (which may be collapsed onto a 

The position of the point C, therefore, is arbitrary and A, B, 
E, D are the vertices of a parallelogram. 

Problem 54. Given are two lines a, b and the three points P, Q 
and R. Construct a path consisting of three line segments 
which has the properties: 

(a) the path starts on a and ends on b, 

(b) P, Q and R are the mid-points of the three segments. 

Problem 53. Five half-turns satisfy the relation 

H e HqH c H b H a = H c . 

What geometric condition on the disposition of the points is 
characterized by this relation? 
We first rearrange the relation: 

Ff E H a HcH B H A = H c o H c H e H d H c H b H a = I. 


We now study the set ft of all isometrics, that is, of all transfor- 
mations which preserve congruence, ft is the set of all finite 
products of reflections: 

Q = M t M t ...Mr M { . 

£5 is a direct (orientation preserving) or an opposite (orientation 
reversing) isometry according to whether n is even or odd. 

Problem 55. Show that the set ft under the operation of com- 
bination of transformations is a group. 

ft is called the group of plane isometrics. 

We have caUed figures congruent if they can be mapped onto 
each other by mappings in the group ft. This notion of con- 
gruence has the following remarkable properties. 

1. Every figure is congruent to itself: F t s F t (property of 

2. Iff, is congruent to F 2t then F 2 is congruent to F t and vice- 
versa : 

F t St F 2 o F 2 = F, (property of symmetry). 

3. If jF t is congruent to F 2 and F 2 is congruent to F 3 , then F, is 
congruent to F 3 : 

F, s F 2 and F 2 s F 3 s> F x at F 3 (property of transitivity). 

Definition. The properties of reflexivity, symmetry and transiti- 
vity characterize an equivalence relation. 




Equivalent figures form what is known as an equivalence 

As examples of equivalence classes with respect to con- 
gruence, we have: 

the set of all squares with sides of a given length d; 

the set of all pairs of points a distance d apart; 

the set of all right-angled triangles with sides 3 in., 4 in., 5 in. 

Problem 56. Show that the concept of parallelism defined on 
p. 28 is an equivalence relation. 

We now want to give a complete description of the mappings 
in ft. We first consider the direct isometries and show: 

Theorem 10. A direct isometry with two distinct fixed points is 
necessarily the identity transformation /. 




Assume that 4> has the two fixed 
points Fi and F 2 (see Figure 48). 

Then the line/ joining F, to F 2 is — pf 

obviously a poinlwise fixed line. 
Now let P be an arbitrary point in 
the plane and h be the perpendicular 
from P to /. Since the right-angle 
between / and h is left unchanged 
by the mapping $, we conclude 
that h is also a fixed line. There Fig. 49 

are now two possibilities to con- 
sider: either P' = P or P' and P are symmetrically placed with 
respect to /. This latter possibility is ruled out since * is an 
orientation preserving isometry. Hence, all points of the plane 
are fixed points, that is, <J> = 1. 

Now take an arbitrary direct isometry Q, that is, a product of 
an even number of reflections. Let A'B' be the image under il of 
a line segment AB. Problem 34 tells us that there are two re- 
flections M p and M q such that M q M p maps AB onto A'B' (see 
Figure 34). Hence, M p M q Q is a direct isometry with the two 


fixed points A and B and is, therefore, the identity. However, 

MpMqQ = /<*>£1 = M q M r 

We state this as: 

Theorem 11. Every direct isometry can be expressed as the 
product of two reflections and is therefore a translation or a 
rotation. The direct isometries are the elements of SB. SB is 
known as the group of direct isometries. 











Fig. 49 

Now we turn to the opposite isometries. We start with a 
simple example, the mapping 

* = M h M„M f , 

where/ and h are perpendiculars to g (see Figure 49). 

Remembering that the order in which two reflections are 
performed can be interchanged if their axes are orthogonal, we 
deduce that 

<fc = M h M g M t = M h M f M B = M t M h M P 

But M h M f = 7" is a translation; hence we can write 

<!> = TM g = M a T. 

Our orientation reversing isometry is, therefore, the product 
of a reflection and a translation with a vector v parallel to the 

axis of reflection. We call this type of opposite isometry a 
giide-refiection. For v = this reduces to a reflection. Reflec- 
tions are, therefore, special cases of glide-reflections. 

Theorem 12, Every opposite isometry is a glide-reflection. 



Fig. 50 

In order to prove this surprising fact we consider an opposite 
isometry Q. Let A'W be the image of a line segment AB under 
Q (see Figure 50). 

According to Problem 34, we can find two lines p and q such 
that M q M p also maps AB onto A'B'. We now introduce the 
line r which joins A' and B'. We easily verify that M p M q M t Cl is 
a direct isometry with the two fixed points A and B. From 
Theorem 10, it must be the identity /: 

M p M q M r Q = /• 

Q - M r M q M p . 

Hence, £2 can be written as a product of three reflections. It 
remains to show that every product of three reflections is a 

We have to distinguish between the two cases: 

(a) p, q, r are parallel. In this situation it is easy to see that £1 
can be reduced to a single reflection. 

(b) p and r are not both parallel to q. Let us assume that q and 
r intersect in a point R. Then M r M^ represents a rotation which, 
according to Theorem 7, can also be written as a product of 



Fig. 51 

two different reflections. Wc choose s perpendicular to p and a 
line h such that M r M q = M h M t . We then have 

Q = M r M*M p m M„M s M p = M h H s . 

Now split the half-turn H s into two reflections M s and M 3 , 
where /is parallel to, and g perpendicular to, ft. This gives us a 

fi = M h H s = M„M t M p 

which can immediately be seen to be a glide-reflection with the 
axis g. 
Now we can state: 

Theorem 13. The group ft contains only four types of mappings, 

namely translations, rotations, reflections and glide-reflections. 

Note that Theorem 13 also tells us that every mapping Q of 

ft can be represented as a product of at most three reflections. 

Problem 57. What type of glide-reflection is CI = M r M q M f , if: 
(a) p, q, r form an equilateral triangle ; 
(b)p, g, r form a right-angled triangle; 
(c) p, q, r all pass through the point S? 

It is to be noted that a glide-reflection with a fixed point is a 



Problem 58. Show that, if the lines p, g and r form a triangle, 
then the axis of the glide-reflection £2 = M,M q M 9 passes 
through two of the feet of the altitudes of the triangle. 

Problem 59. Two congruent line segments AB and A'B' are 
given. Construct the axis, g, and the vector, v, of the glide- 
reflection which maps A onto A' and B onto B'. 

One way of finding g and p is to express £1 as a product of 
three reflections as in Figure 50, and to reduce this to another 
triple product as in Figure 49. There is, however, a simpler way. 
If we choose the line / in the representation H = M h M t M f 
= H H Mf, given in Figure 49, to pass through A, then A is a 
fixed point of M } . Hence A' is the image of A under H s . We 
deduce that the mid-point of AA' must lie on the axis g (see 
Figure 52). Using this property we can immediately construct 
g from the two pairs of points A,A' and B, B'. 






\ A 1 

Fig. 52 

Problem 60. Investigate the following isometries: 

(a) M t H Sl if S lies on g ; 

(b) M g H St if S does not lie on g ; 

(c) H A M f H A ; 
<d) M„H F M . 

Discuss the decomposition of half-turns into suitable products 
of reflections. 

Problem 61. Verify the following equivalence: 
(M g H s ) 2 = /■**> S lies on g. 



The group St contains two types of involutions: reflections 
and half-turns. These simple mappings bear a close relation to 
the basic elements of plane geometry: lines and points. Corres- 
ponding to every point and every line there is a unique trans- 
formation. A geometric relation of incidence between points 
and lines corresponds to a group relation between the corres- 
ponding transformations. For example in Problem 61 we found 
the equivalence 

1 . M S H S M S H S = / *> S lies on g. 
Another example of equivalent statements is 

(F, C, //, A'are the vertices of a paral- 

2. H K H H H G H F = lo jlelogram (which may be collapsed 

lonto a line). (See Figure 22.) 

We now wish to look for other equivalences of this kind. 

(M t M q M p ) 2 = I means that the opposite isometry M^M^M,, 
is a reflection M f ; hence M^M F m M r M f . The mapping which 
is here described in two ways is either a rotation or a translation. 
In the first case p, q and r all pass through a point O, in the 
second case the three lines are parallel. In either case p, q, r lie 
in a pencil of lines. 

We have, therefore, the equivalence 

3. {M r M q M^ 2 = Iop,q, r belong to a pencil. 

If F is the mid-point of the line segment AB, then H F H A and 
H B H F both represent the same translation. Hence 

4. H F H B H r H A m lo Fis the mid-point of AB. 
Problem 62. Verify the following equivalences: 

5. M f H B MjH A = /-*>/is the mediator of AB; 



6. M f M b M f M a = / of is a bisector (or mid-parallel) of a 
and b; 

7. (H F H B T(H F H A r = I\ fFdivid. 

(m,n positive integers)] jternally 

divides the segment AB in- 
in the ratio n:m; 

8. (H B H F )"(H F H A ) m = l\ <s> fF divides the segment AB ex- 
(/n,« positive integers)] <: * jternally in the ratio mm; 

9. H B M f H B H A M s H A = lofis parallel to the line AB; 

10. (H B MfH A ) 2 — lof is perpendicular to the line AB; 

11. H S H C H S H B H S H A = /■**• S is the centroid of the triangle 

All the group relations on the left-band side have the same 
structure; they are products of transformations representing 
the identity. We shall call such products cycles of reflections and 

These equivalences enable us to translate geometric relations 
concerning position into algebraic language, and to translate 
back. This idea leads to a new method of proof for geometric 
theorems. The following problem serves to introduce this 

Problem 63. Show that a line 
through the mid-points of two 
sides of a triangle is parallel to 
the third side. 

We begin the proof by trans- 
lating the given geometric re- 
lations of position into algebraic 

Fig- 53 

Fis the mid-point of AB o H F H B H F H A = /; 
G is the mid-point of AC o H G H C H G H A =* /. 


From these we deduce 

— HqHfHgHaHa,Hf 

= (H G H F ) . 

(using Theorem 4) 

The result H C H B = (H G H F ) 2 means that BC ■ 2FG. We 
have proved even more than was required, for we have also 
shown that the segment FG is half as long as BC* 

Here we have proved a theorem in geometry by means of 
algebraic manipulations with half-turns. This method can be 
extended and other groups of mappings can be used to provide 
proofs of geometric theorems. 

Problem 64. Prove, using reflections and half-turns, that the 
mediators of the sides of a triangle arc concurrent. 

We restrict ourselves here to giving an outline of the solution. 
Let /and g be the mediators of the sides AB and AC, and let a 
be the line joining D, the point of intersection off and g, to the 
vertex A. Then M g M a M f is obviously a reflection in a line. It is 
then easy to show that the axis of this reflection is the third 

We conclude our study of the group of isometries with the 
following remarks. It is now possible, using the knowledge of 
mappings and groups which we have gained so far, to define 
the somewhat vague idea of geometry more exactly. Geometry, 
in the sense of Felix Klein, is a study of those quantities and 
properties of figures which are left invariant under all the 
mappings of a group; it is the study of invariants of a certain 
group of mappings. For every group of mappings we can, 
therefore, develop a particular geometry. 

Corresponding to the group ft of isometries is the so-called 
geometry of congruences, which formed the main part of the 
traditional elementary geometry course. 

* It should be noted that this is a special case of the theorem that rays 
through a given point are divided by parallel lines in equal ratios. No 
assumptions about continuity arc required for our proof of this special 


The group ft contains translations, rotations, reflections and 


We now wish to enumerate all the iso- 
metries which map a square ABCD onto 
itself. Obviously, no translation or glide- 
reflection can map the square onto itself, 
and so wc need only consider rotations and 
reflections. It is easy to see that the square 
is mapped onto itself by the eight isometries 


a=e' shown in Figure 54. 

Identity I 

-9 9 

C-A' D=A> 
-0 Q- 

A=0' BzA' AzC' B=D'A=B ! S=C' 

Rotation Rotation Rotation 

through* SO' t hrough +100' through* 270 ' 
ff, ff z R 3 

D=D' C=A' D=B' C*C's D=A' C=B' D*C> , C=0' 


b=b' a=a' b=d' a=d' b=c' a=b' b*a' 



M 3 

Reflections about the axes shown 


M t 

Fig. 54 



There are no other isometries which leave the square invariant 
and so the product of any two of these mappings must again be 
one of the eight. For the same reason, the inverse of each of 
these mappings must also belong to this set of eight. Therefore, 
we have a group which has only a finite number of elements. 
We have already met finite groups defined by means of numbers 
and so it should not surprise us to find that finite groups also 
occur in transformation geometry. 

Before discussing this group further we wish to know how we 
can find the product of any two of its elements. This will con- 
siderably simplify the working out of the group table. The map- 
ping R 2 > f° r example, maps A onto C, B onto D, C onto A and 
D onto B. We express this by writing 


(A B C D\ 

* 2 = \c DA b\- 

(A B C D\ 

3 ~ \d c b a y 

B C 
C B A 
For the mapping M 3 R 2 (first R 2t then M 3 ) we have 

A B C D 

R 2 I i i i 

C D A B 

M 3 i i, I I 

B A D C 

that is, 


* " (j 

I C £>j 

[b a d cy 

This, however, is the mapping M A . Hence, we get the relation 

M 3 R 2 = M A . 

If, in this way, we work out all the products we get the 
following table.* 

* Note that wc find the product M 3 R 3 by looking in the row labelled M 3 
and the column headed /?.. 





/ R t 



M x M 2 Mi Af 4 


I Rr 



| M t M 2 Mi M A 


Ri Ri 



1 M 3 M A M 2 M x 

R 2 

R 2 R 3 



] M 2 M t M A M 2 

R 3 

Ri I 



1 A/« M 3 Mi M 2 

M x 

M % M 4 

M 2 


'/ R 2 Ri Rt 

M 2 

M 2 M 2 

M v 

M 4 

\R 2 I Ri Ri 


Mi M t 

M A 


,Ri R 3 I R 2 

M A 

M A M z 


M t 

| ^3 -Ki R2 * 


This group table has the following remarkable properties. 

In every row and every column, every element (mapping) 
occurs exactly once. This is a property of all group tables. 

I and the elements R t , R 2 and R 3 form a group by them- 
selves (see the top left-hand quarter square of the group table). 
That is, our group of eight elements contains a subgroup of 
four elements consisting of those mappings which preserve the 
orientation of the square. Here again we learn a general fact 
about groups of mappings : if we demand additional invariants, 
then the group is restricted to a subgroup. 

The group is non-commutative. It does contain pairs of 
elements which commute, but we have, for example, 

M 2 R 3 = M A and R 3 M 2 = M 3 . 

Since group multiplication is associative, we can work out 
higher products. This corresponds geometrically to the product 
of more than two mappings. For example. 

MiR 3 M 2 R 3 M A = (M \R 3 )M 2 (R 3 M A ) 

= M 3 (M 2 M 2 ) 
= M 3 I 
= M 3 . 

Every other way of working out this product leads to this 
same result. This is the content of the associative law, that is, 
the group postulate IV. 

Hence, we get the same result by the following bracketings of 
the factors: 





M 1 R 3 M 2 R 3 M 4 = (M 1 R 3 )(M 2 R 3 )M 4 = (M 3 M 4 )Af 4 = R 2 M A 

= M 3 ; 
M 1 R i M i R s M 4 , m A/,i? 3 Af 2 (/? 3 A/ 4 ) - M l R 3 {M 2 M 2 ) 

= M t R 2 = M 3 . 

We stress once more that the operation of combining trans- 
formations is always associative. 

We shall now demonstrate a general theorem of group theory 
by considering our finite group. Denote our group of isometries 
by © 8 . We can see immediately that the two elements / and R 2 
form a subgroup § 2 . If we now multiply both elements of & 2 by 
an element of <5 8 not belonging to § 2 — say by M t — we obtain 
the set of elements 

M& 2 = {M t I, A/,/?,} = {M u M 2 }. 

These are two elements of (5 8 not belonging to § 2 « If we take 
another clement of <§ 8 which has not occurred so far, for 
example, M 3 , we obtain in the same way 

M& 2 = {M 2 l, M 3 R 2 ) - {M 2 , M 4 }. 

If we repeat this process with one of the remaining elements 
of <5 8 , for example, R lt we obtain 

*i$2-{*iA*i*a} = {*i,* 3 }. 

Now we have a complete partitioning of the elements of (<5 8 
based on the subgroup |> 2 . We may speak of it as a decomposi- 
tion of (5 8 with respect to § 2 , and we write 

62 = &* + {Mi, M 2 } + {M 3I Mt} + {R U R 9 ). 

The elements of <5 8 by which we 'multiplied 1 § 2 are not 
uniquely determined. We remark, however, without proof that 
the decomposition of © 8 with respect to §» 2 is, nevertheless, 
unique ; we always obtain the same four classes. The essential 
fact is that <5 8 can be divided into a number of classes which all 
contain the same number of elements, namely the same number 
of elements as are contained in § 2 . This is true for every sub- 
group § of a finite group (5. After a finite number of steps the 
process of splitting into classes terminates and we always 

obtain a number of classes with the same number of elements in 
each of them. So we have verified in a special case the general 

If a finite group © has a subgroup §, then the number of 
elements in § is always a divisor of the number of elements in 

Our group © 8 can, therefore, only contain non-trivial sub- 
groups with 2 or 4 elements. The following is a complete 
enumeration of the subgroups of © 8 . 

{/, R 2 }, {I, A/,}, {/, M 2 }, {I, M 2 }, {I, M 4 }, 
{/, R u R 2 , R 2 }, {I, R 2 , M lt M 2 ], {I, R 2 , M s , M A ). 

The subgroup {/, M x } is characterized geometrically by the 
fact that the vertices B and D are fixed points under mappings 
of this group.* Try to characterize the remaining subgroups of 
<5 S in this way. 

According to the particular field of mathematics which has 
given rise to the group in question, this theorem contains a 
number theoretic, algebraic or geometric statement— the group 
concept reveals common mathematical structures. 

We now wish to connect our group © 8 with a group of 
permutations. Here we touch upon the field of mathematics 
where the idea of a group first appeared. 

If we only consider the four vertices A, B, C, D of our square, 
every isometry in © 8 produces a one-one mapping of the point 
set {A, B, C, D} onto itself. We have already mentioned on 
p. 19 that such one-one mappings of a finite point set onto 
itself are called permutations.! 

Our group of isometries which leave the square invariant 
consist of permutations which preserve relations of proximity. 
That is, in the square the vertices A and C are adjoining B; this 
property is preserved by mappings of © 8 . There are other 

• This is another example of the previously mentioned fact that the choice 
of a subgroup corresponds to an increase in the invariants. Now not only 
is the square as a whole invariant but also two vertices slay fixed. 

t Often the actual result of the mapping, that is, a particular order of the 
elements, is called a permutation. 


permutations of {A, B, C, D} apart from the eight in <5 8 . For 
example, the permutation 

v-i A B C D \ 
\b da cj 

does not occur in (5 8 ; after this mapping C and D are the ad- 
joining points to B (we must think of the cyclic order of the 
four points). Altogether there are as many permutations of four 
elements as the number of different orders of these elements. 
This number is =4! = 24, 

for we have four possible choices for the first place, for the 
second place there are three choices, for the third place we have 
two elements to choose from, and the remaining element goes 
into the fourth place. The 24 permutations of the four elements 
A, B, C, D can be characterized by the second line of our 
permutation symbol; in alphabetical order they are: 

ABCD (r) 



DABC (r) 







CBAD (w) 



BCDA (r) 





CD AS (r) 





DCBA (m) 

These permutations form a group of which our group of 
mappings, © s , is a subgroup. We verify again in this example 
that the number of elements in a subgroup is a divisor of the 
number of elements in the full group; 8 is a divisor of 24. In 
the table of the 24 permutations, the rotations are denoted by 
(r) and the reflections by (m). 

Problem 65. Find the group table of the isometries which map 
an equilateral triangle onto itself. 

Problem 66. The two permutations 

1 ~{C H D F B A G E ) 



and ¥, = < y 


are elements of the group of permutations of eight objects. 

(a) Determine the permutations *Pi -1 , x ¥ 2 ~ l , *F 2 2 . 

(b) Determine the permutations ^^^ ^j^ and V^ 2 '¥ t ~ 1 . 

(c) Show that H? 2 is an involution; 

(d) Which is the smallest exponent p for which *iEV = /? 

(e) Show that the elements 4\, «F, 2 , ^j 3 , . . . , T^ -1 , ¥^ = / 
form a group. 

For the solution of 
(d) consider a set of 
eight points and indi- 
cate the mapping de- 
fined by Tj (see Figure 
55). From the cycles 
appearing in this figure 
we can easily compute p. Fig.55 

Problem 67. Consider each capital letter of the alphabet in 
turn and find the group of isometries which map the figure onto 
itself. Divide the letters into classes by putting all letters having 
the same group of isometries into the same class. 





Fig 56 


BO 69 BS 83 


Problem 68. Repeat Problem 67 for the frieze patterns shown in 
Figure 56 which we imagine to be extended indefinitely in both 

Problem 69. Verify the relation 

(OT)" 1 =*P- 1 *'" 1 , 

which holds for all groups, and give some examples taken 
from the groups we have considered. 



Fig. 57 

Let 5 be a fixed point in the plane and ft a non-zero number 
(positive or negative). We define a mapping of the plane by the 
following rules: 

A point A and its image A' lie on the same straight line through 
S, that is, A, A' and S are collinear. 

Moreover, each pair A, A' satisfy the relation 


-=-r = p (here we think of directed line segments). 

This mapping is called an enlargement, S is the centre and /i . 
the scale factor of the enlargement. If n>0, the corresponding 
points A and A' lie on the same side of S. If, however, ^<0, 
then S lies between A and A'. An enlargement with scale factor 
- 1 is the same as a half-turn about S. 

If we are given S, A and A', then it is easy to construct the 
image of any other point B (see Figure 57). We draw the line 
through A' parallel to AB to intersect SB in B'. 
We then have, 

SB' SA' 
SB ~ SA ~ **' 



It is just as easy to find the original point B if we are given the 
image point B". The construction shows that enlargement is a 
one-one mapping. In general, S is the only fixed point of the 

Here is a list of other properties of enlargements which will be 
used in later constructions. 

From Figure 57 we can see that if a point P moves on the 
lineg, only the ray of projection SP' will change in the drawing. 
In particular, the parallel g' remains fixed. This means that if P 
moves along g, its image P' will describe the line g';g' is, there- 
fore, the image of g. 

1. Enlargements are line-preserving transformations. More- 
over, a line g and its image g' are always parallel. 

We note that lines through S are mapped onto themselves, 
although not all their points stay in the same place. They are 
fixed lines but not pointwise fixed. 

As each pair g, g' and h, K of corresponding lines is parallel, 
the angle between g and h is preserved by the transformation ; 
in other words: 

2. Enlargements are angle-preserving transformations. 

We have, further, the relation 


S A1 


= /*> 

which means: 

3. The lengths of corresponding line segments are in the ratio 

If fi>0, corresponding line segments are parallel in the same 
sense; if #<0 they are parallel but are in the opposite sense. 
We deduce immediately the following property: 

4. The ratio of the lengths of two image line segments is equal 
to the ratio of the lengths of the original line segments. 
Thus, enlargements preserve ratios of distances. 



In a triangle and its image under enlargement the altitudes 
are also corresponding line segments, hence the areas are in the 
ratio 1 :/?. Since polygons can be divided into triangles, and 
figures with a curved boundary can be approximated arbit- 
rarily closely by means of polygons, we have: 

5. The areas of a figure and its image under an enlargement are 
in the ratio 1 ip 2 . 

Further, we have: 

6. The inverse transformation to an enlargement E(S,n) is the 

enlargement £" '(5, l//i). 

We shall now look at a series of problems which can be solved 
by enlargements applied to a geometric locus. 

Fig. 58 

Problem 70. Construct a line through one point of intersection, 
S, of two circles T, and F 2 , such that the two chords on this 
line have lengths in the ratio 2:3. 

Let X and X' be the other points of intersection of the 
required line with the circles I\ and F 2 (see Figure 58). Then 
we can write 

SX' _3 

SX 2' 



Hence, X' is the point of intersection of T 2 with the image of F t 
under the enlargement with centre 5 and scale factor — 3/2. 

The enlargement with scale factor 3/2 gives another solution ; 
in this case one of the chords lies on the other. 

Problem 71. Given are two squares with a common vertex A. 
Draw a line through A such that the lengths of the segments 
inside the squares are in the ratio 1 :2. 

Problem 72. Given are two circles T^ and r 2 , and a point S. 
Construct a point A on Fl and a point B on T 2 such that S is the 
mid-point of AB. 

Problem 73. Construct the triangle ABC, given the lengths of the 
sides AB and AC, and the length of the bisector of LB AC. 

Let AB = c, AC = b and let the length of the angle bisector 
A W be w. We begin the construction by drawing A W. A geo- 
metric locus for B is the circle F with centre A and radius c 
(see Figure 59). Now we know that W divides BC in the ratio 
c:b. An enlargement with centre W and scale factor -b[c will, 
therefore, map F onto a geometric locus for the vertex C. On 
the other hand, C must lie on the circle with centre A and radius 



Problem 74, Construct a triangle ABC in which BC is of length 
a, the median from B is of length m, and the lengths of the sides 
AC and BA are in the ratio 1 :X. 

We start by drawing the line segment BC. The Apollonius 
circle T defined by BP:PC = X:l is then a geometric locus for 
A. An enlargement with centre C and scale factor i maps T~ 
into a geometric locus for the mid-point A' of AC. However, 
A' also lies on the circle with centre B and radius m. 

Problem 75. Construct a triangle ABC given the lengths of the 
sides AC and AB, and the length s of the median from A. 

We begin the construction by drawing AB = c. The circle T 
with centre A and radius b {— AC) is then a geometric locus for 
C. This can be transformed into a locus for C\ the mid-point 
of BC, by an enlargement with centre B and scale factor ■£. 
Another locus for C is given by s. 

If we begin the construction by drawing the line segment 
AC, then the problem can be solved by using a half-turn. 

In the following examples enlargements are used in an 
essentially different way. The common method of solution is by 
relaxing one, or possibly more, of the conditions. This is done in 
such a way that the remaining conditions determine a family 
of figures which can be mapped onto each other by enlarge- 
ments. After constructing one member of this family, we trans- 
form it by an enlargement so that its image satisfies the omitted 
conditions as well. Thus, the scale factor of the enlargement 
used depends on the choice of the representative, that is, upon 
the way we draw the initial figure. 



Problem 76. Given a triangle ABC, construct a square with two 
vertices on BC and one vertex on each of the sides AB and AC. 

If we ignore the condition that J should lie on AB, a family of 
squares is determined. The square R'S'T'U' (see Figure 60) is 
one representative of this family. By a suitable enlargement 
with centre C, we can map 7" onto T without disturbing the 
other incidences. The scale factor of this enlargement depends 
upon the choice of R'S'T'U'. 

We leave it to the reader to show that the problem could also 
be solved using enlargements with centre A and with centre B. 

Problem 77. Draw a circle to pass through two given points A 
and B and to intersect the given line s in a chord which sub- 
tends an angle of 120° at the centre of the circle. 

The centre M of the circle to be constructed lies on m, the 
mediator of AB (see Figure 61). We first consider the family of 
all circles with centres on m which satisfy the condition on their 
intersection with s. This is a family of circles which can be 
mapped onto one another by enlargements with centre S, the 
point of intersection of m and s. By a suitable mapping of a 
representative circle T' we can obtain a circle T which satisfies 
all the given conditions. Note that the problem has two solu- 
tions, since there are two points A t ' and A' 2 on T' which 
can be mapped onto A. Figure 61 shows only one solution. 

Fig 6i 


We now investigate the transformation obtained by com- 
bining two enlargements. Assume that: 


E t is an enlargement with centre S t and scale factor ft Y ; 
E 2 is an enlargement with centre S 2 and scale factor n 2 . 


Fig- 62 

A point A is mapped onto A t by E lt and A t is mapped onto 
A' by E z (see Figure 62), The transformation E 2 E l leaves the 
line s through S x and 5 2 fixed, since s is a fixed line for both 
enlargements. The image of an arbitrary line h through A is 
parallel to h and through A'. From this it follows immediately 
that the line g, joining A to A', is another fixed line of E 2 E lt 
The point S where the lines g and s meet is, therefore, a fixed 
point of E 2 E X . We now want to calculate the position of S on s. 
In order to do this we introduce the following vectors: 

SiS 2 = rf, S 2 S' = p (S' is the image of S under E{), 
SiS = x t S 2 S = q. 

Then we have the following relations : 

p+d = HiX, q = fi 2 p, d+q = x. 

From these we obtain 

x = d+q = d+ft 2 p = rf-t-AftiLX-O = (l-^d+tt^x. 

Solving this vector equation with respect to x, we obtain 

x - £=&_ d * 

I-H,fi 2 

* The position of S relative to S, and 5, can also be represented by the 
fraction X = f£. We find A = *~^ . 



This shows that 5 does not depend upon the choice of A ; S is 
determined by the two mappings E l and £ 2 . For the transforma- 
tion E 2 E X , therefore, corresponding pairs of points A, A' lie 
on lines through the fixed point S. 

From Figure 62 we can also recognize the following relation : 

SA' _ S'A l SA'_ 

SA ~ SA 'S'At V^ 2 ' 

Hence, we have shown that E 2 E t is an enlargement with 
centre S and scale factor « = PiH 2 . 

The product of two enlargements E x and E 2 is, in general, 
again an enlargement, E. The three centres S lt S 2 and S are 
collinear, and the scale factors satisfy the relation ft = fixfi 2 . 

In Figure 63 this result is illustrated by the transformation of 
the triangle ABC. A new group of mappings begins to be recog- 

Fig 63 

Problem 78. Given are two enlargements E^S^) and 
E 2 (S 2 , -2/3). Construct and calculate the position of S, the 
centre of the mapping E 2 E l . 



To construct S, map an arbitrary point A (which does not 
lie on the line SiS 2 ) by £ 2 £i- The position of S can be calculated 
by the vector formula given above. Here we have 

l-(-2/3) 5 
I-2<-2/3) a 7 

Problem 79. Find the product of the two enlargements £[(5, ,2) 
and E 2 (S 2 ,i). 

If we apply the transformation E 2 E X to a point A we obtain a 
situation like that shown in Figure 64. It can be seen that for 
every pair of corresponding points A and A', the vector from A 
to A' is v = id. Hence, E 2 E t is obviously a translation with the 
vector v. 

Since, for our pair of enlarge- 
ments, fi x fi 2 = 1, the denominator 
in our formula for x vanishes. If 
the value of /i^ is ver Y c l° se t0 
1, then the vector x will be very 
long. We can take fi l fi 2 = 1 to 
characterize the point at infinity 

on the line through S, and 

s * f i 2 Fig. $4 

We can now consider translations to be special cases of 
enlargements ; the centre of enlargement for a translation is a 
point at infinity and the scale factor is 1. 

Problem 80, Two enlargements £ t and E 2 are given with 
/ii/t 2 = 1. Show that v, the vector of the translation E 2 E U is 
given by 


» = (i-£>-a-i'j* 

Problem 81. T is a translation with vector v and E an enlarge- 
ment with centre S and scale factor /i#l, Show that ET and 
TE are enlargements with scale factor ft and construct the 
centres of these transformations. 

• The point at infinity is not a proper point. We consider it as an ideal 
element corresponding to the value X = 1 for the ratio of distances of a 
point from two given points. 



We can use the same method as in the case of the two finite 
enlargements. The line through S parallel to v is a fixed line of 
TE = £' and ET = £". In Figure 65 the centres S' of £' and 
S" of E" have been obtained by finding the images of a point 
A. It is easy to verify that 

x < = — I „ a nd x" = 2-rir. 

fi — 1 ft — 1 

If we use the equivalence 

£' = 7£ <=>£'£"' =7", 

and the corresponding one for E", then these relations can be 
deduced from the result of Problem 80. 

Theorem 14. The set of all enlargements forms a group, 91, 
under the operation of combination of transformations. 

To prove this we must show that the group postulates are all 
satisfied. When we consider the product of two elements of 21 
we have to consider the possibility of the transformations 
being translations. 

The unit element of 21 is the identity /. As in the case of 
rotations, / can be represented in different ways; every enlarge- 
ment with ft m 1 represents / provided the centre of enlarge- 
ment is a finite point. 

The group is not commutative; this means that, in general, a 
product depends on the order of the factors. We give as an 
example of two elements which do not commute, the two 
transformations J and £ of Problem 81. There are, of course, 


commutative pairs in 21 ; for instance, two enlargements with 
the same centre always commute. 

Problem 82, Show that a pair of enlargements, neither of which 
is a translation, commute if, and only if, both mappings have 
the same centre. 

Two enlargements commute if E 2 E X and E X E 2 have the same 
centre. This is obviously true if S x m S 2 , that is, if d — 0. 

Now assume that d ^ 0. The centres of E 2 E X and E X E Z can 
then be described by the two vectors 

x > = l ~J}K d and x » = -±Zl±-d. 

l-^iJ'2 l-JHth 

x' and x" describe the same point if x'~ x" = d (see Figure 
66), that is, if 

1-^1^2 l-A»iA*2 
This equation can be transformed to 

(l-^)(l-,t 2 ) = Q. 

But this can never hold since n Y # I and fi 2 ^ I. 

If p x n 2 = If it can easily be deduced from the formula in 
Problem 80 that the two translations E 2 E X and E X E 2 have 
different vectors (unless lii+fi 2 = 2, in which case {i L = (i 2 
= 1). 

Fig, 66 



Problem 83. An enlargement E x with centre S t maps A onto A v . 
A second enlargement E 2 with centre S 2 maps A x onto A^'. 
Hence, A x is the image of A under the transformation E 2 E l . 
Construct the image of A under the transformation E X E 2 . 
We give the solution without comment in Figure 67. 

Problem 84. If £, and E 2 are two enlargements, show that 

EyE^E^y 1 = E^E.-'E^ 1 

is always a translation T with a vector v which has the direction 
of the line joining S x and S 2 . 
This is illustrated in Figure 67; Tmaps the point A x onto 

a 2 : 

Try to express the translation vector of T in terms of fi u [t 2 
and d = S X S 2 . 

Problem 85, Two enlargements E X (S X , /*,) and E 2 {S 2 , ;< 2 ) are 
given. Construct the centre of the mapping E t E 2 E l '. 

Problem 86, Construct the centre of the transformation EiE 2 E u 
where E lt E 2 and E 3 are three given enlargements, and illustrate 
geometrically by means of this figure that the associative law 
holds in St, that is 

E 2 (E 2 E t ) = (E 3 E 2 )E X . 

What incidences can be derived from this? 

Problem 87. Prove the following theorem of Mcnelaus: If a 
line s intersects the sides of a triangle A, A', A* in the points 
Sj, S 2 , S 2 , then 

o [ si & 2 A ^3^ , 

'S i ~A"S 2 ~A r "S^A* ~ 

To prove this we introduce the following three enlargements : 

*(** = S$H** = Wr) E ( s ^ = !£} 

The enlargements with centres on a line s form a subgroup of 
9J; this is a simple consequence of the rules for composition. 


We shall denote this subgroup by 9t s . 

Then E 3 E 2 E 1 c 9tj. But A, which does not lie on s, is a fixed 
point of this mapping, This can only happen if the mapping is 
the identity. However, 

£ 3 £ 2 £, = I ^Hrfhl** - U 
which proves our theorem. 

Problem 88. Prove the following theorem of Desargues: If the 
lines joining corresponding vertices of the triangles ABC and 
A'B'C are parallel, then the points of intersection of cor- 
responding sides lie on a line s. 

The proof, like that of Problem 87, can be reduced to the 
composition of transformations in 91. 



Given two circles T t and T 2 there are, in general, two enlarge- 
ments Ei and E 2 which map oue onto the other. The centres 

Si and S 2 lie on the line through the centres of the circles;* 
the easiest way of finding them is by using parallel radii of the 
circles (see Figure 69). The corresponding scale factors are 

**2 j r i 

}ii=*~r and /*2 = _ r- 

Problem 89. Show that the six centres of enlargement associated 
with the three circles T u T lt r 3 , lie on four lines, three on each 
line. (Theorem of Monge.) 

We pick out two circles r f and F ; . There are two enlargements 
£,j and E lt ' which map T, onto T } ; let E l} be the one with a 
positive scale factor. We immediately deduce the following 

£23^2 = £i3> £23^2 = £13 > 

£23 £l2 = ^13 > ^23 £l2 = £l3> 

which imply the statement to be proved. 

Problem 90. There are two enlargements £ and £* which map 
the circle T onto the given circle T', Describe the mapping 
£*£ _1 . 

The following four problems provide a method, based on 
transformation geometry, for proving Feuerbach's Theorem on 
the so-called nine-point circle. 

Problem 91. Show that, in every triangle, the point of intersec- 
tion of the altitudes H (the orthocentre), the centroid G, and 
the circumcentre O lie on a line (the Euler line). 

Hint: use the enlargement which maps the given triangle 
ABC onto the triangle A'B'C formed by the mid-points of the 
sides of ABC. 

Problem 92. Show that the circumcentre O' of the triangle 
A'B'C lies on the Euler line of the triangle ABC. 

Problem 93. Denote the circumcircles of ABC and A'B'C by 
r and T' respectively. They can be mapped onto each other 
by two enlargements: 

* S, and Sj are often referred to as centres of similitude. 


E(G,-i) and E*(G*&. 
Show that the centre of enlargement C* is H. 

Problem 94. The image of the triangle ABC under E* is a 
triangle A*B*C*. A*, B* and C* are the mid-points of the 
lines joining H to the vertices A, B and C. 

Note that (A', A*), (B', 5*) and (C, C*) are pairs of corres- 
ponding points under the transformation E*E~ l and use this 
to prove that the circle F passes through the feet of the altitudes 
of the triangle ABC. 

The circle I"' has now been shown to contain the mid-points 
of the sides, the feet of the altitudes, and the mid-points of the 
lines joining the orthocentre to the three vertices— r' is Feuer- 
bach's nine-point circle. 


We have already mentioned that: 

The enlargements with centres on a fixed line g form a sub- 
group 9t, of 91, The transformations of 9l fl are distinguished by 
the fact that they all leave the line g invariant. 

If two enlargements have the same centre G, that is, if d = 0, 
then x = 0. Hence, the centre of the product is G also. We 
conclude that the transformations of 91 with a given centre G 
also form a subgroup 9f c . Its elements are characterized by the 
in variance of the point G. If C lies on g, then 

that is, 9I C is a subgroup of 9t r 

We noted earlier that subgroups can be obtained by demand- 
ing additional invariants. 9t s is obtained from 91 by demanding 
that the arbitrary lineg is invariant, and 9t G by demanding that 
the point G is invariant. 

We can also demand the invariance of a whole figure. We 
may, for example, imagine the square and circle pattern shown 
in Figure 70 to be continued indefinitely in both directions. This 
leads to a figure which is invariant under enlargements with 
centre S and scale factor (^/2)" where n is any integer. These 
mappings form a subgroup 9t s ' of % s and therefore of 9t itself. 
Unlike the previous examples 91/ is a discrete subgroup of 91. 


By this we mean that the transformations in 9t s ' do not form 
a continuous family. 

The most general enlargements leaving Figure 70 invariant 
have scale factors ±(-j2)' 1 where n is any integer. These map- 
pings form a group 9( s " and we have the chain of inclusions: 

9l s 'c9I s "c9r s cfc. 

All these subgroups of 91 have infinitely many elements. 





Fig, 71 

Problem 95. A point A and lines b and c are given. Construct a 
square with A as a vertex such that the vertex B lies on b and 
the vertex Conc.C is the vertex opposite A.* 

To solve this we first disregard the condition that C should 
lie on c. We then get a family of squares of which AXX'D* is a 
representative (see Figure 71). In this we can think of X' as the 
image of Sunder a rotation about A through 45°, followed by an 
enlargement with centre A and scale factor ,/2. If X' moves on 
the line b', the image of b under these two transformations, then 
C is the intersection of b' and c. The rotation with = —45" 
leads to a second solution of the problem. 

This method of solution is based on transforming a geo- 
metric locus ; we have used this method before. What is new is 
that here two different types of transformations have been 

• Cf. Problem 23 in which similar conditions were given. 


combined — the transformation used was the product of a 
rotation and an enlargement having the same centre. It is easy 
to see that the two mappings commute; it does not matter 
whether the rotation or the enlargement is performed first. 

We call the product of a rotation and an enlargement with 
the same centre S a spiral similarity. 

This defines a new type of transformation. A spiral similarity 
is characterized by the centre S, the angle of rotation 6, and the 
scale factor // ; we may denote it by iV(S,6,ft). Before we turn 
to the theory of spiral similarities we give two further construc- 
tion problems of the same type as Problem 95. 

Problem 96. Given are a point A and two lines b and c. Con- 
struct a triangle ABC with B on b, C on c, £_BAC = 60° and 
LABC = 45°. 

A procedure analogous to that used in Problem 95 leads to a 
spiral similarity W(A, +60°, sin 45°/sin 75°). 

Problem 97. Given are a point A, a circle T and a line g. Con- 
struct a rectangle with sides in the ratio 1 :2, which has one 
vertex at A, the vertex B on T and the vertex C (opposite to A) 
on g. AB is to be one of the shorter sides of the rectangle. 
The solution to this problem uses a spiral similarity 

W(A,± tan" 1 2, JS). 

The angle of rotation and the scale factor can be constructed in 
a model figure. 

Problem 98. Given are four points A, B, C, D. Find a rectangle 
with sides in the ratio 1 :2, such that each side (or its extension) 
passes through one of the given points. 

This is a generalization of Problem 45. Now the strip bounded 
by a and c is mapped onto the strip bounded by b and d (see 
Figure 72) by the spiral similarity 

W(M, 90°, 2). 

The vector u is mapped onto a vector w which is orthogonal 
to u and twice as long. The translation T with vector t maps the 
image C (= W(C)) onto D. If we draw w with end-point D, 



then the starting point A* is a point of the side b of our rect- 
angle. A* and B together determine b and it is then easy to 
complete the rectangle. 

In solving Problem 98 we had to use the product of a spiral 
similarity and a translation. This means we multiplied an 
enlargement by a rotation and a translation. 

Definition. A finite product of isometries and enlargements is 
called a similarity. 

Similarities, like isometries, can be divided into two classes, 
direct (orientation preserving) similarities and opposite (orienta- 
tion reversing) similarities. 


The spiral similarity is an example of a direct similarity. If 


R(S,0) is a rotation and E(S,p) is an enlargement, then 

ER= RE = W(S,8,p) 

is a spiral similarity with centre S, angle of rotation 0, and scale 
factor fi. 

Fig 73 

For spiral similarities we have the relation 

W(S,6,p) = W(S,0 +« -fi). 

In order to obtain a unique representation, we shall always 
assume that fi>0 in what follows. 

Fig 74 

Problem 99. Let K be a direct isometry and E an enlargement 
with centre S and scale factor p. # 1. Show that the direct 
similarity SI = EK always has a fixed point F. 



We first assume that K is a rotation with centre O through 
angle (see Figure 74), We can always find a point F which is 
mapped onto the same image F' by both K and £"'. To do this 
we take two corresponding points A and A' (= E'^A)) and 
draw the isosceles triangle with angle AO*A' = 0. The triangle 
AO*A' is mapped onto FOF' by a spiral similarity with centre 
S. F is now a fixed point of the transformation EK. 

If K is a translation, then a fixed point can be constructed in 
an equally simple way. 

The definition of similarities may lead us to expect to find 
many different types of transformations in the class of direct 
similarities. Surprisingly, this is not so; we can show: 

Theorem 15. Every direct similarity is a spiral similarity. 

The proof depends on the fact that a direct similarity is 
completely determined by two points A and B, and their images 
A' and B'. 

If the line segment A'B' is congruent to AB, then Q, the 
transformation mapping AB onto A'B', is a translation or a 
rotation. Both are special cases of spiral similarities. 

We now assume that AB and A'B' are not congruent. Let/> be 
the mediator of AA'. The reflection M p maps A onto A'; we 
denote the image of B under this mapping by B*. We now 
introduce g, the bisector of the angle between A'B* and A'B'. 
The reflection M q leaves A' fixed and maps B* onto fit. If we 
denote by E the enlargement with centre A' and scale factor 
H = B'A'jB-\A', we obviously have 

fi - EM q M p = EK. 

AT is a translation or rotation according to whether p and q are 
parallel or not (see Figure 75). 

We now have the situation of Problem 99. We know, there- 
fore, that the mapping Q. has a fixed point F. The isometry 
K m M q M p can also be obtained as the product of two different 
reflections; if we choose />, to be the line through O and F we 
have to use as g { the perpendicular through to the line s 



joining A' and F,f In order not to overcomplicate Figure 75, 
these extensions are shown in Figure 76. We now have 

Q = EM^M, = EM,M Pi = E(M q MXM s M Pt ). 

The introduction of M s 2 = / does not, of course, change the 
transformation. M s M Pi is a rotation with centre F and M q M s 
represents a half-turn, H L , which wc can also interpret as an 
enlargement. Front the rule for combining enlargements, we 
know that EH L is an enlargement with its centre on s. But F 
and fare corresponding points for both E and H L ; hence EH L 
leaves the point F fixed. Therefore, F is the centre of enlarge- 
ment of EH L = E*. 
Thus we have obtained a representation 

a = E*(F,-pi)R(F,4>) = WW,-fi) = WPAft 
Q is a spiral similarity with centre F. 

Problem 100, A, A' and B, B' are corresponding pairs of points 
for the spiral similarity 

Determine the centre S, the angle of rotation 0, and the scale 
factor fi. 

t In the case of a translation, p v is the perpendicular to s through F. 



Fig, 77 

8 appears as the angle between corresponding lines g and g', 
and n = A'B'jAB. 

If A and A' are corresponding points, then /_ASA' - 0. 
Using this property we can find S by the construction indicated 
in Figure 77. 

The product of two direct similarities is again a direct simi- 
larity. If we consider translations and enlargements as special 
cases of spiral similarities, then we can state our result as fol- 

Theorem 16. The spiral similarities form a group S under the 
operation of combination of transformations. 

The group ® contains as subgroups, the group % of enlarge- 
ments, the group ft of isomctries and the group X of transla- 

Problem 101, Discuss the construction considered in Problem 
100 for special positions of the points A, A' and B, B\ 


Problem 102. Given are two spiral similarities 

W&Sift&d and W 2 (S 2 ,B 2 ,n 2 ). 

Determine the centre, angle of rotation, and scale factor of the 
spiral similarity W = W 2 W±. 

To solve this problem we construct two pairs of correspond- 
ing points for W; the problem then reduces to Problem 100. 

The construction is particularly simple if we select A to be 
the point S x and B to be the point which is mapped onto S 2 by 
the transformation W t . We have: 8 =* 0i + 2 i /* = /'i^- 

Problem 103. Construct a cyclic quadrilateral given the lengths 
of the four sides. 

Consider a spiral similarity 
with centre A which maps B onto 
D. The image C of C lies on 
the line through C and D, since 
opposite angles of the quad- 
rangle are supplementary (see 
Figure 78). For this spiral simi- 
larity we have 8 = a and n = dja. 
If we start the construction by 
drawing side c, we can first find 
C, using b' = b.dja. 

We now have two geometric 
loci for A; we have AD = d, 
and also AC: AC = d:a (Apoi- 
lonius's circle). Ft'g7& 

Problem 104. A B is a given line segment and g an arbitrary line 
through A. If we map AB by all possible spiral similarities with 
a given centre S which leave the image of A on the line g, show 
that the geometric locus of B' is a line s. What relation exists 
between the two lines g and 5? 

Consider the different images of AB and interpret the corres- 
pondence between starting point and end-point of these line 
segments as a mapping. 









S Fig 79 

As the simplest example of an opposite similarity we consider 
the so-called stretch-reflection. We obtain this type of transla- 
tion by combining a reflection M s with an enlargement E which 
has its centre on s. As can be deduced immediately from Figure 
79, the stretch-reflection Z does not depend upon the order in 
which the reflection and the enlargement are carried out: 

Z = EM, m M S E. 

The class of opposite similarities is also dominated by a 
single type of transformation: 

Theorem 17. Every opposite similarity is a stretch-reflection. 

The proof is similar to the corresponding one for direct 
similarities. We refer again to Figure 75 and the transforma- 
tions defined there. If, in addition, we denote the line through 
A' and B' by r, we can represent fi* by the following product: 

CI* - EM F M„M P = EK*. 

K* is an opposite isometry. For the further reduction of Ci* we 
consider Figure 80 which illustrates the position of the three 
lines p, q, r and the centre S. 



Fig 30 

From this figure we deduce 

£1* = EM r M q M p m EM ri {M q2 M p ) m M r EH L . 

The transformations M rj and E commute since 5 lies on r 2 , 
EH L m E* is again an enlargement with centre S* on q z . If we 
now introduce the perpendicular 5 to r 2 through S*, we obtain 

Cl* = M, 2 EH L = M f E* m (M ri MXM s E*) - H M E*M S . 

Here we have used the fact that E* and M s commute since 5* 
lies on s. Ef = H M E* is an enlargement with centre S* on s. 
We have now obtained the following representation for CI*'. 

Cl* = EW S . 
This is a stretch-reflection, since S* lies on s. 

The essential elements of a stretch-reflection are the axis of 
reflection s t the centre of enlargement 5 on s, and the scale 
factor fi. 

Problem 105. The two pairs of points A, A' and B, B' are given. 
Describe the stretch-reflection Z which maps A onto A' and B 
onto B'. 

The scale factor n is given by A'B'jAB. It can be seen from 
Figure 79 that the axis s divides the line segment A A' in the 
ration 1 : ft. Therefore we can immediately construct two points 
of 5. 



Problem 106. Show that the set of all spiral similarities and 
stretch-reflections under the operation of combination of 
transformations forms a group ©*. 

©* is the group of all similarities. The geometry of this group, 
in the sense of Klein, is called the geometry of similarity. 

To this geometry belong all those geometric properties of 
figures which are invariants with respect to ©*. Examples are 
the orthocentre of a triarjgle, the mediator as the geometric 
locus of all points equidistant from two given points, the circle 
as the geometric locus of all points equidistant from a given 
point. We can say that circles form an invariant class of curves 
with respect to ©*. Pythagoras's Theorem also belongs to the 
geometry of similarity, for if we have a 1 +b 2 = c 2 for a triangle, 
the same is true for the image triangle. 

Problem 107. The spiral similarities form a subgroup S of <3*. 
Why do the stretch-reflections fail to form a subgroup ? 

Problem 108. Determine the group <5 f of all similarities leaving 
the pattern in Figure 70 invariant. List also some subgroups of 

Problem 109. What kind of transformation is *¥K<¥- ■ if Ke ft 


What do we obtain when K is : 

(a) a reflection, 

(b) a half-turn, 

(c) a translation, 

(d) a rotation, 

(e) a glide-reflection ? 

Definition, If a geometric figure F can be mapped onto a figure 
F' by a transformation *P e S*, then we say that F and F' are 
similar figures. 

Problem 110. Show that the relation of similarity defined above 
is an equivalence relation. 

Problem 111. A triangle ABC is given. Corresponding to each 



side construct a vector which is at right-angles to the side, is 
twice as long as the side, and which points outwards from the 
triangle. Prove that the vector sum of the three vectors is zero. 

Problem 112. A, B, C, D are the vertices of a square (described 
in an anticlockwise direction). Consider the three spiral 

WM> 45°, V2). 

W 2 (B, 45°, 72), 

W 3 {C, 90°, i). 

What transformation is W i W % W x 1 

Problem 113. What kind of transformation is *F 2 if *F is a 



The transformations we have discussed so far were really 
elementary geometry in a new guise. Now we come to the 
first example of a mapping which does not belong to the frame- 
work of elementary geometry. We enter at the same time a field 
of geometry of rather more recent origin; affine transformations 
were first introduced in 1748 by Leonhard Euler (1707-1783) 
in his analysis of the infinite.* 

Let us select a line s in the plane and a direction given by the 
angle a it makes with s. We also choose a real number p, 
positive or negative but not zero. 

Wc can now define a mapping of the plane by the following 
law of correspondence: 

The lines through pairs of corresponding points P, P' are 
parallel to the given direction. 

* Euler L. : Introductio in analysin infinitorum. Tomus secundus. Cap 
XVIII, De similitudine et affinitate linearum curvarum. Opera omnia 1/9. 



For every pair of points P, P' we have 
P'P* _ 

where P* is the intersection of the line PP' with s. 

This transformation is called a perspective affinity^; s is the 
axis of affinity, the given direction is called the direction of 
affinity, and p. is called the scale factor of the affinity. 

When we wish to describe the transformation fully we shall 
use the symbol <b(s, «, p). 

An important special case of a perspective affinity is reflection 
in a line, for 

M E = ®(s, 90", - 1). 

Perspective affinities for which a — 90" are called normal 


If we are given a point P and its image P' under a perspective 
affinity with axis s, then further pairs of corresponding points 
can easily be constructed. We can use the construction shown 
in Figure 81 to ensure that 



- v- 

If Q moves along the line g, then Q' moves along g' (in our 
construction only the parallel to the direction of affinity moves). 
Hence, g' is the image of g under the affinity. 

If A is parallel to j, then the image points of h lie on a parallel 
line h'. This is a consequence of the fact that if P moves along 
the line h, then the distance PP* remains constant; hence 
P'P* is constant, too. 

1. A perspective affinity is a line-preserving transformation. 
Corresponding lines intersect on the axis of affinity. 

The line j is a fixed line; in fact, all points off are fixed points. 
The lines parallel to the direction of affinity are also fixed lines, 
but, unlike s, they are not pointwise fixed. 

t Or axial stretching. 



The verification of the following properties offers no parti- 
cular difficulties' accordingly no detailed explanation is neces- 

2. A perspective affinity maps parallel lines onto parallel lines. 

3. The transformation preserves ratios of division; the ratios of 
distances of three points on a line g are equal to the corres- 
ponding ratios for the image points on g\ 

4. The ratio of the area of a polygon to its image under a 
perspective affinity is 1 : ft* 

5. A perspective affinity is a one-one mapping. The inverse 
affinity has the same axis and direction but has scale factor 

If a perspective affinity is to preserve the (non-oriented) area 
we must have ft = ±1. If ft = - 1 we obtain the so-called 
oblique reflections (see Figure 82(a)), which we can see to be a 
generalization of reflection in a line. The transformations with 
ft = 1 obviously include the identity /; however, this is not the 
only affinity with this property. If we note that the scale factor 
can also be expressed in terms of the oriented distances of 
corresponding points A and A' from s, we deduce that if ft = I 
these distances are equal. If * ^ /, then the direction of affinity 
must be parallel to s. In this case we speak of a shear with the 
axis s (see Figure 82(b)). 

•We are here considering orientated areas. The ratio of the areas is 
positive or negative according to whether the original and image polygons 
have the same orientation or not. 



If « = 0, ft can no longer be chosen arbitrarily, ft is then 
always + 1. For a shear or and /« are not independent, and as a 
consequence the transformation is not completely determined 
by s, <x and fi. A shear is a degenerate type of perspective 

As parallelism is invariant, a square is mapped by a perspec- 
tive affinity onto a parallelogram. A given quadrangle can only 
be mapped onto a square by a perspective affinity if its opposite 
sides are parallel. The following problem shows that this condi- 
tion is not only necessary but is also sufficient. 

Fig. 83 

Problem 114. Find a perspective affinity with given axis s which 
maps a given parallelogram A BCD onto a square. 

The image of a parallelogram under a perspective affinity is 
always another parallelogram, since parallel lines have parallel 
images. Hence, the parallelograms form an invariant class of 
quadrangles with respect to the perspective affinities. The 
image parallelogram is a square if the angle 0' between two 
sides is 90° and the angle <p' between a side and a diagonal is 
45°. The conditions 

e -» 0' = 90° and <£->0' = 45° 

define two geometric loci for A'. Their intersection is A'. Once 
we have a pair of corresponding points we can, of course, easily 
construct the transformation. The complete geometric loci have 



a second poiot of intersection which is the image of A' utider 
reflection in s. If we ask for the transformation, the problem 
has two solutions. The two squares obtained are, however, 
congruent and we shall later prove this. 

Among the numerous properties which characterize quad- 
rangles, only those which characterize parallelograms are 
invariant under perspective affinities. Examples are the proper- 
ties that opposite sides are parallel, or that the diagonals bisect 
each other. In general, a perspective affinity will not preserve 

Problem 115. Show that a hexagon can be mapped by a pers- 
pective affinity onto a regular hexagon if and only if its diagonals 
are concurrent and pairs of opposite sides of the hexagon are 
parallel to a diagonal. 

The two conditions are certainly necessary, for they are satis- 
fied by a regular hexagon and they are invariant under perspec- 
tive affinities. 

We can obtain a hexagon satisfying these conditions in the 



following way (see Figure 84). We start with the three diagonals 
a, b, c intersecting in a point M. We choose a point i>, on a and 
draw through it a line parallel to c; this determines P 2 on b. 
Then draw a line parallel to a to obtain P 3 , etc. Finally we 
obtain P 6 . The parallel to b through P 6 passes through P x and 
completes the hexagon. This follows from the relation 

P t P 2 = MPs = P A P S = MP 6 . 

We can now prescribe the axis of affinity s in an arbitrary 

If we determine the affinity in such a way that a, b, c are 
mapped onto three lines intersecting each other at angles of 
60°, then the image of P t P 2 . . . P 6 will be a regular hexagon, 
since the six triangles which make up the image hexagon will be 

Problem 116. Given a triangle ABC, a line s and a point S\ 
construct a perspective affinity having s as its axis and such that 
S' is the centroid of the triangle A'B'C. 

The position of the centroid of a triangle is determined by 
certain ratios of line segments. Since such ratios are invariants 
under perspective affinities, it follows that the image of the 
centroid of ABC is the centroid of the image triangle A'B'C. 
The notion of a centroid is invariant under perspective affinities. 

We can construct the centroid S of the given triangle ABC. 
S and S' are then corresponding points which determine the 

The centroid S of a quadrangle 
ABCD can be found by two 
different decompositions into 
triangles (see Figure 85). This 
combined construction is also 
affine, that is, it is built up using 
properties which are all invari- 
ant under perspective affinities. 
The affine image of the con- 
struction coincides with the 
corresponding construction for 
the affine image of the quad- 



Problem 117. Verify the con- 
struction of F. Wittenbauer* for 
the centroid of a quadrangle. 
This construction is shown in 
Figure 86 and is also invariant 
under affine transformations. 
The sides of the quadrangle are 
divided into three equal parts 
and a parallelogram is then ob- 
tained by drawing lines through 
pairs of points of trisection ad- 
joining a vertex. The sides of this 
parallelogram are parallel to 
the diagonals of the given quad- 
rangle and the diagonals of the parallelogram intersect in the 
centroid of the quadrangle. 

Wittenbauer's construction can be reduced to the construc- 
tion shown in Figure 85 by purely geometric means, that is, 
without calculation. 

Problem 118. Give a construction for finding the centroid of a 

Problem 119. Given are the axis of a perspective affinity and a 
pair of corresponding points A and A'. Construct two orthogonal 
lines (i and v which intersect at A and which have orthogonal 

The problem is solved by constructing a circle which passes 
through A and A' and has its centre M on s. M lies on the media- 
tor m of the line segment A A'. Figure 87 is based on a perspec- 
tive affinity having a scale factor /*>0. 

Since a perspective affinity preserves parallelism, there are 
two orthogonal directions which are always mapped onto 
orthogonal directions. We always obtain this same pair of 
directions whatever pair of corresponding points we use. We 
call these two directions the invariant right-angle pair of the 

* Sec Strubccker, K., Einfiihriirig in die hshere Mathematik. Vol. ], p. 
403, Munich, 1956. 



Fig 87 

This construction runs into difficulties if the direction of 
affinity is at right-angles to the axis, that is, for normal affinities. 

Consider first the case p 96 — 1, that is, the transformation is 
not a reflection. Then m is parallel to s but different from s (see 
Figure 88). The circle through A A' now becomes a straight tine 
through these points. The invariant right-angle pair exists in 
this case too ; one direction is parallel to, the other orthogonal 
to, the axis of affinity. 

For a reflection (ji = -I) m and s coincide; the point of 
intersection is, therefore, indeterminate (see Figure 89). Every 
point of s can be taken as the centre of a circle passing through 
A and A'. Hence, every right-angle is invariant. The result 



Fig. 89 

Fig. $3 



agrees with the fact that a reflection is an isometry. 

We can sum this up as a further property of our transforma- 

6. Every perspective affinity which is not a reflection has 
exactly one invariant right-angle pair. As far as this property 
is concerned reflections are degenerate cases; under reflec- 
tions every right-angle pair is invariant. 

Problem 120. A perspective affinity is given by its axis s and a 
pair of corresponding points A and A'. Construct an invariant 
60°-pair of lines which meet at A. 

We first consider the case ft>0, that is, A and A' lie on the 
same side of s. Then our problem is to construct a circle through 
A and A' such that the axis s cuts off an arc which subtends an 
angle of 120° (or 240°) at the centre of the circle. This problem 
has already been solved in Problem 77. 

If /*<0, that is, if A and A' lie on opposite sides of s, we first 
reflect A' in s. This leads to a point A" and reduces the problem 
to the case /t>0 already considered. The problem always has 
two solutions. 

In the solution for /i<0 there appears, besides the perspec- 
tive affinity <J>, the reflection A/ s . In order to find the invariant 



60 c -pair for the transformation C>, we first determine the 
corresponding pair for the mapping M&. This is again a per- 
spective affinity with axis s. 

The importance of perspective affinities lies in their connec- 
tion with the parallel projection. For this reason construction 
problems on perspective affinities can be found in many collec- 
tions of problems in descriptive geometry. Therefore we 
restrict ourselves here to considering certain problems which 
bring out new aspects. 


The remark at the end of Problem 120 leads us to the conjec- 
ture that the perspective affinities with a given axis form a 
group. In order to check this we shall use methods of analytic 

We consider first a 
perspective affinity <b(s, 
at, fi) and set up a rect- 
angular coordinate sys- 
tem as shown in Figure 
91. The .v-axis is taken 
to coincide with the axis 
of affinity 5. Our trans- 
formation 0, which 
maps A onto A', is then 
characterized by the 

x' = x+(ft— 1) cot a y, 
/ = 99* 

However, we have seen that shears cannot be characterized 
by the parameters a and u. In order to find a standardized 
representation for all perspective affinities we have to look for 
more suitable parameters. 

Since the transformation O preserves parallelism, the angle 
between a, the perpendicular from A to s, and its image line a' 
is independent of the choice of A. Corresponding to every 
perspective affinity 3> there is a well-defined angle a. Now we 

Fig 9} 




x—x' 1— U 

tan a m ■ > — = £— 

y ft tan a 

Qi— 1) cot a m — ^ tan <r. 

We now replace a in our transformation equations by the 
parameter a. The equations can then be written : 

x' = x— ft tan a y, 
/ = W> 

and it is easy to see that these equations are also valid for 
shears. Tan a always has a finite value.* 

Problem 121. A perspective affinity * has axis s, angle 
a = 45°, and scale factor ft = J. Construct a pair of correspond- 
ing points. 

Problem 122. Calculate the angle a' for the inverse transforma- 
tion <E» -1 {G> is defined as in Problem 121). 

We now have the necessary analytic tools to prove the 
following theorem: 

Theorem 18. The perspective affinities with a given axis s form 
a group Qj. 

Let *i and <t> 2 be two perspective affinities having the same 
axis 5. We use the notation (see Figure 92) 

A(x,y)— >A'(x',y')-^ A"(x",y"). 

Then the equations of the transformations are 

= x'~ft 2 tanff 2 /, 

ft W = x-^ tan &, ,v, jx" = *>-§ 

%' - ft* %" = ft* 

The geometric composition of <t>! and <& 2 so as to form the 
new transformation * = <t> 2 <&[ corresponds to the algebraic 

* Note that n and a are directed angles. 


substitution of the equations for *, into those for * 2 . We 

jc" = x-ti x tan ff, y-Pifo tan a 2 y 
= x—(jii tan ffi +/i,/i 2 tan a 2 )y, 

y" - fhfhy- 

These equations, too, have the structure of transformation 
equations for a perspective affinity with axis s. If we write 

x" = x-t* tan a y, 

y" = ny< 

then we obtain by comparison, 

fi - ftft. 
fi tan a = /*i tan ffi + /i,jt 2 tan a 2 . 

This implies the following rules of composition for the scale 
factors and angles: 

* * * I * m ft*** 

' * I tan tr = l//i 2 tan ffj +tan cr 2 . 

Hence, the mapping obtained by combining two perspective 
affinities with axis s is again a perspective affinity with axis 5. 
The verification of the remaining group postulates is trivial. 

Corresponding to every line 5 there exists a group Q,. This 
fact is expressed by using the suffix s in the symbol for the 



%^ \ 

/ L \ .\J&z 

Fig 92 



The group Si s is not commutative. This can be seen immedi- 
ately from the rule for obtaining the angle of the product 
transformation, since the formula for a is not symmetric in the 
pairs 0*i, tan a % ) and {fi 2 , tan a 2 )- 

Problem 123. Show that in the group Si, we have the following 

rule for obtaining the direction of affinity of the product 
transformation : 

tan a = (1-^ifo) tan ct t tan« 2 

(1 -a*!) tan a 2 +Ml -j»a5 tan %* 

Problem 124. Prove the theorem of Menelaus using transforma- 
tions belonging to the group Q, (cf. Problem 87), 

We are now in a position to settle a question which arose in 
connection with Problem 114. Then we constructed an affinity 
with a given axis s which mapped a parallelogram onto a square. 
We found that there was a second transformation having this 
property and that 

a> 2 = Af,*,. 

But from this it follows at once that the two squares obtained 
are congruent (and are symmetrically placed with respect to the 

Problem 125, Show that two transformations * lf <t> 2 c Q s 

commute if, and only if, their directions of affinity coincide 

(«i = «a)- 
From the rules of composition we found 

1 1 

* 2 *j = *,*2<»— tan 0-,+tan s 2 = — tan <r 2 + tan a t . 
A*2 Hi 

This can be written 

r^; tan °> = t~ ta *>■ 

But the expression on the left-hand side represents cot «j and 
that on the right-hand side cot a 3 , hence, ot, = a 2 , 
We can easily find subgroups of O r We note that s is a fixed 



line for all * e £),. By demanding further invariants we obtain 
subgroups of £5j. 

If we demand that a certain direction should be an invariant, 
we obtain the subgroup O jW of all the transformations having 
the same direction of affinity (given by the angle a). It follows 
from Problem 123 that C 1(a) is a commutative group; the 
composition of mappings in the group being given by the 
relation p = Hiit 2 obtained above. 

Another subgroup Q S '<=C S can be obtained by demanding 
the invariance of the measure of area. O,' contains the perspec- 
tive affinities with ft ■» ±1, that is, shears and oblique reflec- 
tions. The shears alone form a group Q/'eQ',. 

Problem 126. Show that in the group Si" the composition of 
transformations can be described by the relation 

tan g = tan ^H-tan <r 2 . 
Illustrate this result diagrammatically. 


Problem 127. Interpret the two constructions shown in Figure 
93, in which the two image figures have the same area as the 
original figures, in terms of mappings belonging to O,. 



Problem 128. Show that shears having the same axis s and such 
that tan a = na where n is an integer and a a constant, form a 
group. Try to give a geometric figure which is invariant with 
respect to this group (cf. Figure 70). 

The product of perspective affinities having different axes is, 
in general, not a perspective affinity. For example, the product 
of two reflections is a rotation or a translation. These are 
mappings having either one fixed point or none at all and, 
therefore, cannot be perspective affinities. 

Problem 129. Two perspective affinities <&i and <D 2 have axes 
which intersect in a point S, The direction of affinity of each 
mapping is parallel to the axis of the other, and we have 
ft, = p 2 = ft. Show that ^O^ is the enlargement E(S,n) and 
that the two transformations <b x and <b z commute. 

Fig 94 

Figure 94 illustrates the situation. The problem reveals the 
fact that every enlargement E can be represented as a product 
of two perspective affinities. 

Problem 130. Two perspective affinities 4>, and 2 have axes 
jj and s 2 which intersect in a point S. Again we assume that the 
direction of affinity of each mapping is parallel to the axis of 
the other, but now we suppose that //,/x 2 = 1. 

(a) Show that <tf 2 Q> l is a transformation which preserves lines 
and area, and that S is a fixed point of the transformation. 

(b) Let A be a point on s t and B a point on s 2 . Show that the 



lines through the pairs A, B' (= <J> 2 *i(£)) and B, A' 
(= 9 2 9x(A)) are parallel. 
(c) Show that <I> 2 <t>, = 4>i1> 2 . 

Problem 131. <t>! and <P 2 are two given perspective affinities 
satisfying the conditions of Problem 130. Setting fi t = A, we 
see that ft 2 = \fX, and that we can write 

^O, = *(A). 

(a) Show that the transformations 4>(A) (for all X # 0) with 
given axes 5! and s 2 form a commutative group &(s u s 2 ). 

(b) Translate the group relation 

W.) = WJMd 

into geometric language. (Use the result of Problem 130(b) — the 
Pappus-Pascal configuration.) 

Problem 132. Show that hyperbolae with asymptotes 8% 
and s 2 are invariant under transformations of the group 

If *i and <D 2 are two arbitrary perspective affinities, then the 
transformation * 2 *i * s obviously a mapping which preserves 
lines, parallelism and ratios of division. These properties of 
<!>! and * 2 are not lost by composition. We have met other 
transformations with these properties, for example, the ele- 
ments of the groups ft and S*. 

Definition. Every finite product of perspective affinities is called 
a general affine transformation or, for short, an affinity. 

It follows from the definition that affinities are transforma- 
tions which preserve lines, parallelism and the ratios of division. 
They form a group Q, the full affine group. The groups £Jj are 
subgroups of Q. Moreover, ft and S* are also subgroups of Q, 
since we have seen that the reflections, which generate ft, are 
particular perspective affinities, and Problem 129 showed that 

t This property is used in the geometric introduction of the natural 
logarithm . See, for example, Van der Waerden B. L., Die Einfiihrung des 
Logarithmus im Schulumerricht. El. Math. 12, 1957. 



every enlargement could be obtained as the product of two 
perspective affinities. 

Problem 133. Given are two triangles ABC and A'B'C Show 
that there exist a perspective affinity <t» and a spiral similarity 
Q such that the transformation £1$ maps the triangle ABC onto 
the triangle A'B'C 

Take, for example, the spiral similarity which maps the line 
segment AB onto the segment A'B'. Then it is easy to find a 
perspective affinity * such that £1* has the required property. 
The reader will easily work this out with the aid of a simple 

The transformation QO which we have just constructed is an 
affinity. Hence, there always exists an affinity mapping a given 
triangle ABC onto a second given triangle A'B'C, and there 
exists only one such affinity as we shall proceed to show. If we 
want to construct the image of an arbitrary point P, we can do 
this in the following way. We draw lines through P parallel to 
AC and AB. Let these lines intersect AB and AC in the points 
P t and P 2 respectively. Since ratios of division are invariant 
we can immediately construct the images i>,' and P 2 ' on the 
sides A'B' and A'C of the image triangle. If we then draw 
lines parallel to A'B' and A'C through these points, we find 
P', the image point of P. Hence, the image of any point, and 
therefore the affinity, is uniquely determined by the triangles 
ABC and A'B'C. There can be only one such affinity. The 
decomposition into a spiral similarity and a perspective affinity 
is, however, not unique. 

The group Q has finite subgroups as well. If we ask, for 
example, for the affine transformations which leave a given 
triangle ABC invariant, then we see that any such transforma- 
tion produces a permutation of the three vertices. Since there 
are only six permutations, it follows that there are only six 
affine transformations of the triangle onto itself, and that these 
six elements form a group. We shall follow up this example by 
looking in more detail at the affinities which map a given 
parallelogram A BCD onto itself. Every such affinity again 
results in a permutation of the vertices. Since the affinity is 
determined by the images A', B\ C of the points A, B, C, we 



can characterize the affinity by giving the positions of A', B', 
C on the parallelo- 
gram. This choice is -^o \ c, 

not arbitrary, for D has 
to be mapped onto a 
vertex as well. This will 
only happen if A', B', C 
are (in this order) ad- 
joining vertices of the 
parallelogram. There 
are eight possibilities 

for this. Four of these preserve the orientation, and four re- 
verse the orientation of the parallelogram. There are, therefore, 
eight affinities which leave the parallelogram invariant. They 
form a group Q 8 and the eight mappings can be described as 
follows (see Figure 95): 

/ identity transformation 

"t>, affinity moving the vertices round one place in an anti- 
clockwise direction 

<D 2 affinity moving the vertices round two places in an anti- 
clockwise direction 

4> 3 affinity moving the vertices round three places in an anti- 
clockwise direction 

$/ oblique reflection with axis s, 

<I> 2 ' oblique reflection with axis s z 

d> 3 ' oblique reflection with axis s 3 

4> 4 ' oblique reflection with axis s 4 . 

The group table for these transformations is 



<K 2 

* 3 


* S ' 



















<J> 2 ' 




a> 3 















0> 2 ' 































4» 3 ' 








If we compare this table with the group table for the iso- 
metrics which map a given square onto itself, we see that the 
two groups have the same structure although they originated 
from different problems. We can establish a one-one corres- 
pondence between the elements of the group <5 8 for the square 
and the elements of S2 8 , in such a way that the product of two 
elements of © 8 corresponds to the product of corresponding 
elements in Q 8 . Two groups for which such a one- one corres- 
pondence can be defined are said to be isomorphic. 

In Problem 114 we showed that every parallelogram can be 
mapped onto a square by some affinity. If we denote this 
affinity by T, then we have the following relations between the 
elements of <5 8 and Q 8 : 

Rj = "¥$^-1 

and we have, for example, 

Mj = T*/^ 1 , 

M 3 R t =(¥it> 3 'T- 1 )(V<Di4'- 1 ) 

This describes the isomorphism between the two groups 
geometrically. The two problems leading to the groups <5 8 and 
Q 8 were not, therefore, basically different. The transformations 
4>i, * 2 , $3 are rotations deformed by an affinity, and *,', 
®i, GV. */ are reflections which have been deformed. 

Another interpretation of the isomorphism springs from the 
fact that the groups © 8 and Q 8 describe the same group of 
permutations of the vertices A, B, C, D. 

Problem 134. Show that the perspective affinities with the same 
direction of affinity form a group. 

Problem 135. Show that if an affinity has a pointwise fixed line 
s, then it is a perspective affinity with axis s. 

Let A, A' be a pair of corresponding points for the given 
affinity <J> and suppose that A does not lie on s. Making use of 
the fact that affinities preserve ratios of division, we can con- 
struct the image point P' of any point P from s, A and A'. This 
image is unique; hence it follows that there is only one affine 
transformation which leaves all points of s fixed and maps A 



onto A', But the perspective affinity with axis 5 and the pair of 
corresponding points A, A' is such an affinity. 

Fig 9$ 

Problem 136. Show that every perspective affinity <& can be 
represented as a product of a spiral similarity Q. and a normal 
affinity *„. 

Let * be a perspective affinity with the axis s, and let A and 
A' be a pair of corresponding points under 4>. Construct the 
invariant right-angle pair at A and A' (see Figure 96). Now we 
consider a spiral similarity £2 with centre S, angle and scale 
factor ft = SA'fSA. 

The images of A, g and h under the transformation Q are 
denoted by A", g" and h". By the definition of fl we have 
A' = A", g' = g" and h' = h". For the image T" of T under 
fi, we have 

A'T" m nAT. 

Now let <&„ be the normal affinity with axis s a = g' = g", 
which maps T" onto T, Then the transformation 0„£1 has both 
S and Tas fixed points. Since ratios are invariant it follows that 
every point on the line j through S and T is also a fixed point. 
Hence, using the result of Problem 135, $„Q is a perspective 
affinity with axis s. However, it maps A onto A' and so it must 




be the given affinity fl>. Hence, <& = d>„£i which shows that any 
perspective affinity can be decomposed into a spiral similarity 
and a normal affinity, f 


Affine geometry, in the sense of Klein, is the study of invari- 
ants of the group G. Hence, all those properties of figures which 
are preserved under affine transformations belong to affine 

The group @*, generated by reflections and enlargements, is 
a subgroup of Q. This means that every property belonging to 
affine geometry also belongs to the geometry of similarities. 
The reverse is not true; a property of similarity geometry is not 
necessarily an affine invariant. For example, we mentioned 
before that the relation 

a 2 + b 2 = c 2 , 

for the sides of a triangle, is an invariant under similarities. An 
affine transformation will not, in general, preserve right-angles. 
Since this relation characterizes right-angles it will not, there- 
fore, be an invariant under Q. Affine geometry embraces only 
a subset of the properties which belong to similarity geometry. 
The theorem of Pythagoras is not one of these; the group of 
similarities forms an extreme limit for this particular theorem. 
However, ratios of areas are conserved by affine transforma- 
tions. To measure an area means to find out how many times a 
certain unit area is contained in it. The measure of area is, 
therefore, based on a relation of the form 

F, = pF 2 . 

From the result of Problem 133 we conclude immediately 
that this relation is preserved by affine transformations. For the 
affine images we have again 

Ft = pjy. 

An example of a property which is affine invariant is that the 
diagonals of a parallelogram divide the figure into four triangles 
of equal area. 

+ This decomposition is not unique. Wc could, for example, start wish a 
spiral similarity with centre 7". 



The notion of a vector also belongs to affine geometry. This 
follows from the fact that parallel line segments having equal 
length are mapped onto image segments which are again paral- 
lel and of equal length. Every relation between vectors is an 
affine invariant; for example, if 

v m a+3A, 

then die same relation holds for the image vectors. 

Every group defines an equivalence relation between those 
figures that can be mapped onto each other by transformations 
of the group. For example, all triangles are equivalent with 
respect to S, for an affine transformation can be found which 
maps any given triangle onto any other triangle. There is only 
one class of triangles in affine geometry. In similarity geometry, 
however, such triangles as the isosceles right-angled triangles 
form a class by themselves. Parallelograms also form one class 
in affine geometry, but not trapezia. Two trapezia only belong 
to the same equivalence class if the ratio of the lengths of the 
parallel sides is the same. Only then can one be obtained from 
the other by an affine transformation. 

Problem 137. What conditions must be satisfied by a hexagon 
if it is to belong to the same equivalence class as a given regular 
hexagon : 

(a) with respect to the group ®, 

(b) with respect to the group Q? 

Problem 138. Show that two quadrangles belong to the same 
equivalence class with respect to Q if the diagonals in both 
figures divide each other in the same ratio. 

Problem 139. Show that areas of corresponding polygons under 
a general affine transformation are in a fixed ratio ft. 




We start with the following definition : 

Definition. The image of a circle under a perspective affinity is 
called an ellipse. 

Since the perspective affinity preserves ratios, every diameter 
of the circle T is mapped onto a diameter of the ellipse V. This 
fact can be expressed in the language of transformation 
geometry by writing 

<&H„<!>- 1 = H, 


Corresponding to orthogonal diameters of the circle we ob- 
tain what are called conjugate diameters of the ellipse (see 

Fig $7 




Figure 97). The property of orthogonality is not an affine 
invariant, but orthogonal diameters of a circle can be charac- 
terized by a property which is invariant under affine transforma- 
tions. This property is that the tangents at the end-points of one 
diameter are parallel to the other diameter. Conjugate dia- 
meters of an ellipse will, therefore, also be characterized by the 
fact that the tangents at the end-points of one diameter will be 
parallel to the other diameter. Among these pairs of conjugate 
diameters there will be one pair which is orthogonal; this pair 
defines the axes of the ellipse. The axes are, therefore, related to 
the invariant right-angle pair of the affinity. 

Problem 140. Construct the axes of an ellipse which is the image 
of a circle T under the perspective affinity <t>. 

Problem 141. Show that every ellipse is the image under a 
normal affinity of the two circles on its axes. 

We imagine that the given ellipse is the image of a circle T 
under the perspective affinity $. If we construct the invariant 
right-angle pair for at M and M' , then we can reduce the 
problem to the decomposition of $ into a spiral similarity fl 
and a normal affinity <S> ni as described in Problem 136. We 
denote the two normal affinities by ® nt and O rt2 . If a and b are 
the lengths of the semi-axes of the ellipse, we have 

vjsu 90 s , ±^j and <S>Js 2i 90°, ±f\. 

The sign of p t and fi 2 depends on the sign of the scale factor 
of the transformation O. 

Problem 142. Show that $ n 2 _1< Diit ' s an enlargement with 
centre M', and deduce the method shown in Figure 98 for 
constructing points on an ellipse from points of the circles on 
its axes. 

Draw in the tangents to the curves at the three related points 
P u P 2 &n6P\ 

We now wish to show that every ellipse T' can be obtained as 
a plane section of a circular cylinder. According to Problem 141 
there exists a normal affinity <t>„ 2 which maps the circle T 2 on 
the minor axis onto I"'; the scale factor of the affinity being 







I s i I «' 

s 2 

— -^ Figs 

alb. We now imagine the ellipse V turned out of its plane. We 
choose s 2 , the axis of <&„ z , as the axis of rotation and make the 
angle of rotation 4> ~ cos -1 6/dr. The points P 2 and P', which 
were corresponding points under O n2 , will, after this rotation, 
lie on a line orthogonal to the plane of the circle F 2 . Hence, the 
ellipse now lies on the cylinder over the circle F 2 (see Figure 

Fig 99 

Problem 143. Show that the ellipse is the locus of all points 
whose distances from two fixed points F t and F 2 have a constant 



We use the fact that every ellipse can be obtained as the 
intersection of a plane e with a circular cylinder. 

Imagine that we take two spheres of radius b and push them 
into the cylinder from opposite ends until they touch the plane 
e. Let the two points of contact be F, and F 2 . It is then easy to 
show that for all points P' on the ellipse the sum P'F i +P'F 2 
has a constant value. The generating line of the cylinder 
through P' touches the two spheres in points A and B. We now 
use the fact that all tangents from a point P' to a sphere have the 
same length. Hence, we obtain 

P'F l +P'F 1 = P'A+P'B m AB. 

AB is the segment of the generating line through P' which is 
bounded by the two circles of contact between the spheres and 
the cylinder. It can be seen immediately that AB does not 
depend upon the position of the point P' on the ellipse.* 

The two points F t and F z are known as the foci of the ellipse. 

Problem 143 introduced an important metric property of 
the ellipse. Here, however, we are mainly interested in the affine 
properties of ellipses and the following theorem takes us back 
once more to affine geometry. 

Theorem 19. A pair of conjugate diameters determines exactly 
one ellipse. 

The content of this theorem is not trivial since a pair of conju- 
gate diameters can be mapped onto orthogonal line segments 
having equal length, by infinitely many affinities. It is not 
immediately obvious that the inverse transformations will 
always map the circle onto the same ellipse. 

We prove this by giving a construction for an ellipse starting 
with a pair of conjugate diameters. It is necessary that this 
construction is an affine invariant, that is, is built up from 
affine invariant operations such as drawing parallels, or dividing 
line segments in given ratios. We do not allow such operations 
as drawing right-angles. 

Let AB be a diameter of the circle F (see Figure 100), C a 

* This method of proof is due to G. P. Dandelin (1794-1 847). The two 
spheres appearing in the proof are sometimes called Dandeiin's spheres. 



point on F, and q an arbitrary perpendicular to AB. Since C lies 
on the circle, LACB = 90°. We now extend AC until it meets 
q and we letter this point T. His the orthocentre of the triangle 
ABT. The line AP is the third altitude and is, therefore, perpen- 
dicular to BT. We conclude that P, the point of intersection of 
the lines BT and AH, lies on T. If we let q move (still, however, 
remaining perpendicular to AB), then P will describe the whole 
circle. Hence, Figure 100 shows a construction which allows us 
to find any number of points on a circle without using com- 
passes, provided a diameter and one other point of the circle 
are given. Consequently we can solve the following problem 
without using compasses : 

Problem 144. Given two mutually orthogonal diameters AB 
and CD of a circle P, construct the point of intersection of T 
with a line g through B. 

The solution is given in Figure 101(a). 

Apart from drawing the right-angle between q and AB all 
the steps in the construction are affine. We can, however, 
interpret the construction completely in affine invariant terms, 
if we interpret q not as the perpendicular to AB, but as a paral- 
lel co the tangents to the circle at A and B (that is, q is the con- 
jugate direction to AB). It should be mentioned that P can be 
obtained directly by drawing the perpendicular to g through A. 


Fig 107 

However, the affine nature of the construction is lost by this 
short cut. 

An affine translation changes Figure 101(a) into Figure 

We can now base the proof of our uniqueness theorem on 
Figure 101, If a given pair of conjugate diameters A'B' and 
CD' is mapped by a perspective affinity onto an orthogonal 
pair AB, CD which have the same length (the position of the 
axis of affinity is irrelevant), then the inverse mapping applied 
to the circle construction shown in Figure 101(a) always pro- 
duces the same image. The reason for this is that all the points 
and lines constructed in Figure 101(b) depend only on the given 
pair of conjugate diameters. The position of P' on g' is, there- 
fore, uniquely determined. Jf we now rotate g' round the point 
B', then P' describes a well-defined curve — the ellipse with the 
given conjugate diameters.* 

Problem 145. The mid-point of the line segment HT in Figure 
101(a) lies on the tangents to the circle F through C and P. Use 
this property to obtain a construction for drawing tangents to 
an ellipse. 

* The remarkable configuration of Figure 100, on which this proof is 
based, is given in Volume 3 of Enzykfopadie der EkmetUarmathematik by 
Weber-Wellstein. It is a special case of Pascal's Theorem. 



Fig 102 

Problem 146. An ellipse is given by means of a pair of conjugate 
diameters. Determine the points of intersection of the ellipse 
with an arbitrary line g. 

We mention this simple problem on ellipses because its solu- 
tion introduces a new means of using transformation geometry. 

The given ellipse can be mapped onto a circle by an affinity. 
The simplest mapping to use is one which leaves one of the 
conjugate diameters invariant. That is, we consider a perspective 
affinity <f> which has one of the two diameters as its axis of 
affinity. The line g is mapped onto g' by <J>. Now the problem 
has been transformed into one concerning a circle. We have to 
find the points of intersection of g' with the image circle and 
then construct the images of these two points under the map- 
ping O -1 (see Figure 102). 

In this problem we meet an idea which is frequently used in 
higher mathematics. A given problem is reduced by means of a 
suitable transformation to a simpler case for which the solution 
is already known or at least obtainable more easily. The solu- 
tion of the general case is then obtained by considering the 
image of the simpler solution under the inverse transformation. 
This is an extremely useful method of solution. 

The following problem can be solved in the same way. 

Problem 147. Two ellipses are given whose major and minor 
axes are parallel and in the same ratio. Find their points of 



An affinity which transforms one ellipse into a circle will also 
map the second ellipse onto a circle. Hence, a suitable affinity 
reduces the problem to that of finding the points of intersection 
of two circles. 

Problem 148. Construct an ellipse which has the given diameter 
AB and passes through two given points P and Q. 

An ellipse can be constructed if it can be represented as the 
affine image of a circle. Hence, we have to find a perspective 
affinity which maps the required ellipse onto a suitable circle. 

Fig 103 

We take the line AB to be the axis of affinity of our mapping. 
In a circle every chord is orthogonal to the line joining its mid- 
point to the centre of the circle. The equivalent property for an 
ellipse is that a chord and the line joining its mid-point to the 
centre of the ellipse define conjugate directions. Therefore we 
have to choose the affinity in such a way that the angle 4> (see 
Figure 103) is transformed into a right-angle. If we translate the 
angle tf> to P we have, together with the circle with diameter 
AB, two geometric loci fozP' (since P' lies on the circle with UV 
as diameter). Hence, we obtain two solutions for P' — the 
points P t ' and P 2 '. Correspondingly there are two affinities 



* t and <& 2 to be considered. We now want to show that both 
affinities map P onto the same ellipse, that is, there is only one 
ellipse with the required properties. 

We note that P 2 is obtained from JY by the reflection M a . 
For the inverse transformations we therefore have 


= ar 1 */.. 

However, the reflection M s leaves V invariant, hence, the 

image of T' is the same for <b\ 
has a unique solution. 

- 1 

and <t 2 • Thus the problem 

Problem 149. Two ellipses e t and e 2 have a common diameter 
AB. Given P on e t and Q on e 2i and the tangents at P and Q, 
construct the common tangents to the two curves. 

To solve this we map e t and e 2 by two perspective affinities 
onto the circle e' with diameter AB: 

e «- 

<b v <D 2 

Then <& 2 -1 <&i = <P is a perspective affinity which maps ei 
onto e 2 . The direction of affinity of <S> gives the direction of the 
common tangents of e v and e 2 . If we denote the axis of affinity 
of * 1 and <S? 2 by s, then the transformations 

also map e t and e 2 onto the circle e'. Since we have 

this combination produces the same affine relation between 
ei and e 2 . The four possible combinations of 4> lt 0/ and 
*z» ®i produce only two different transformations; to each 
of these corresponds two common tangents of e t and e 2 . 

Problem 150. Show that all ellipses belong to the same equiva- 
lence class with respect to SD. 


The bibliography of the German edition listed 16 books of 
which the following are published in English. 
klein, F., Elementary Mathematics from an Advanced Stand- 
point (Vol. 2, Geometry), Macmillan, New York, 1939. 
weyl, H., Symmetry t Princeton University Press, Princeton, 

An O-level course based on transformation geometry can be 
found in the textbooks of the S.M.P. In Book T the emphasis 
is on construction problems involving a single transformation — 
the transformations considered are the isometries (excluding 
glide-reflection), enlargements and shears. Book T4 introduces 
products of transformations and the glide-reflection. The group 
properties of transformations are not developed but an account 
is given of the description in matrix terms of transformations 
having a fixed point. Similar work on transformation matrices 
is contained in Matthews. The Teacher's Guide to Book T has 
a valuable section on motion geometry which will certainly be of 
great assistance to any teacher who wishes to try out this 
approach to geometry. Further guidance on class-room 
presentation is contained in Chapter 10 of the A.T.M. hand- 
book (this also mentions some of the group applications). 

Rather more advanced accounts of the subject can be found in 
two Russian translations: Yaglom, in which only isometries 
are considered, and Kutuzov. Coxeter in his excellent, compre- 
hensive book gives a short but readable account of the various 
groups of transformations. 

A footnote on p. 43 tells the reader where information on the 
description of patterns and figures in group theoretical terms 
can be found. Terpstra's pamphlet is intended to supplement 
Escher's collection of designs and these, together with the 
books by Weyl and Wells, make fascinating reading. A read- 
able and inexpensive approach to symmetry and plane patterns 
is given by Bell and Fletcher. The work on cycles of half-turns 
and reflections on p. 72 is also done by Thomsen (but this 




article is now of interest mainly from a historical point of view) 
and a full scale axiomatic treatment of this group theoretic 
approach to geometry can he found in Bachmann (German). 

Readers to whom group theory is new should follow up one 
or two of the ideas hinted at in this book. A more detailed 
account of cosets (the name given to the sets appearing in the 
decomposition of © 8 on p. 78) and a proof of Lagrange's 
Theorem (here stated without proof) can be found in any book 
on groups. One of the cheapest and most readable is Leder- 
mann. Problem 109 has deep implications in group theory as 
the reader will discover if he investigates normal and conjugate 
subgroups. An exciting, but expensive, treatment of groups is 
given by Papy. 

a.t.m. handbook, Some Lessons in Mathematics, Cambridge 

University Press, London, 1964. 
bachmann, f., Aufbau der Geometrie aits dem Spiegehtngs- 

begrijff, (Grundlehren der mathematischen Wissenschaften, 

96), Springer, Berlin, 1959. 
bell a. and fletcher T. J., Symmetry Groups, A.T.M. 

Mathematics Teaching Pamphlet No. 12, 1964. 
coxeter, H. s. m., Introduction to Geometry, New York, 

London, Wiley, 1961. 
escher, M. c, The Graphic work of M. C. Escher, Old bourne 

Press, London, 1961. 
scutuzov, B. B., Geometry, (Studies in Mathematics, 4), 

School Mathematics Study Group, New Haven (Conn.) 

ledermann, w., Introduction to the Theory of Finite Groups, 

Oliver and Boyd, Edinburgh, 1949. 
Matthews, G., Matrices (2 vols.), Arnold, London, 1964. 
papy, g., Groups, Macmillan, London, 1964. 
school mathematics project, Book T, Cambridge Uni- 
versity Press, London, 1964. 
school mathematics project, Teacher's Guide for Book 

T, Cambridge University Press, London, 1965. 
school mathematics project, Book T4, Cambridge 

University Press, London, 1965. 
school mathematics project, Teacher's Guide for Book 

T4, Cambridge University Press, London, 1966. 



TERPSTRA, p., Some notes on the Mathematical Background 
of Repetitive Patterns, Oldbourne Press, London, 1961. 

thomsen, g., "The Treatment of Elementary Geometry by a 
Group-calculus", Mathematical Gazette, 17, No. 224 (Octo- 
ber 1933), pp. 230-42. 

wells, A. F., The Third Dimension in Chemistry, Clarendon 
Press, Oxford, 1956. 

YAGLOM, I. m., Geometric Transformations, Random House, 
New York, 1962. 

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